Category: analog Communication

Numerical Problems 2 In the previous chapter, we have discussed the parameters used in Angle modulation. Each parameter has its own formula. By using those formulas, we can find the respective parameter values. In this chapter, let us solve a few problems based on the concept of Frequency Modulation. Problem 1 A sinusoidal modulating waveform of amplitude 5 V and a frequency of 2 KHz is applied to FM generator, which has a frequency sensitivity of 40 Hz/volt. Calculate the frequency deviation, modulation index, and bandwidth. Solution Given, the amplitude of modulating signal, $A_m=5V$ Frequency of modulating signal, $f_m=2 KHz$ Frequency sensitivity, $k_f=40 Hz/volt$ We know the formula for Frequency deviation as $$Delta f=k_f A_m$$ Substitute $k_f$ and $A_m$ values in the above formula. $$Delta f=40 times 5=200Hz$$ Therefore, frequency deviation, $Delta f$ is $200Hz$ The formula for modulation index is $$beta = frac{Delta f}{f_m}$$ Substitute $Delta f$ and $f_m$ values in the above formula. $$beta=frac{200}{2 times 1000}=0.1$$ Here, the value of modulation index, $beta$ is 0.1, which is less than one. Hence, it is Narrow Band FM. The formula for Bandwidth of Narrow Band FM is the same as that of AM wave. $$BW=2f_m$$ Substitute $f_m$ value in the above formula. $$BW=2 times 2K=4KHz$$ Therefore, the bandwidth of Narrow Band FM wave is $4 KHz$. Problem 2 An FM wave is given by $sleft ( t right )=20 cosleft ( 8 pi times10^6t+9 sinleft ( 2 pi times 10^3 t right ) right )$. Calculate the frequency deviation, bandwidth, and power of FM wave. Solution Given, the equation of an FM wave as $$sleft ( t right )=20 cosleft ( 8 pi times10^6t+9 sinleft ( 2 pi times 10^3 t right ) right )$$ We know the standard equation of an FM wave as $$sleft ( t right )=A_c cosleft ( 2 pi f_ct + beta sin left ( 2 pi f_mt right ) right )$$ We will get the following values by comparing the above two equations. Amplitude of the carrier signal, $A_c=20V$ Frequency of the carrier signal, $f_c=4 times 10^6 Hz=4 MHz$ Frequency of the message signal, $f_m=1 times 10^3 Hz = 1KHz$ Modulation index, $beta=9$ Here, the value of modulation index is greater than one. Hence, it is Wide Band FM. We know the formula for modulation index as $$beta=frac {Delta f}{f_m}$$ Rearrange the above equation as follows. $$Delta=beta f_m$$ Substitute $beta$ and $f_m$ values in the above equation. $$Delta=9 times 1K =9 KHz$$ Therefore, frequency deviation, $Delta f$ is $9 KHz$. The formula for Bandwidth of Wide Band FM wave is $$BW=2left ( beta +1 right )f_m$$ Substitute $beta$ and $f_m$ values in the above formula. $$BW=2left ( 9 +1 right )1K=20KHz$$ Therefore, the bandwidth of Wide Band FM wave is $20 KHz$ Formula for power of FM wave is $$P_c= frac{{A_{c}}^{2}}{2R}$$ Assume, $R=1Omega$ and substitute $A_c$ value in the above equation. $$P=frac{left ( 20 right )^2}{2left ( 1 right )}=200W$$ Therefore, the power of FM wave is $200$ watts. Learning working make money

SSBSC Demodulator The process of extracting an original message signal from SSBSC wave is known as detection or demodulation of SSBSC. Coherent detector is used for demodulating SSBSC wave. Coherent Detector Here, the same carrier signal (which is used for generating SSBSC wave) is used to detect the message signal. Hence, this process of detection is called as coherent or synchronous detection. Following is the block diagram of coherent detector. In this process, the message signal can be extracted from SSBSC wave by multiplying it with a carrier, having the same frequency and the phase of the carrier used in SSBSC modulation. The resulting signal is then passed through a Low Pass Filter. The output of this filter is the desired message signal. Consider the following SSBSC wave having a lower sideband. $$sleft ( t right )=frac{A_mA_c}{2} cosleft [ 2 pileft ( f_c-f_m right )t right ]$$ The output of the local oscillator is $$cleft ( t right )=A_c cosleft ( 2 pi f_ct right )$$ From the figure, we can write the output of product modulator as $$vleft ( t right )=sleft ( t right )cleft ( t right )$$ Substitute $sleft ( t right )$ and $cleft ( t right )$ values in the above equation. $$vleft ( t right )=frac{A_mA_c}{2} cos left [ 2 pi left ( f_c-f_m right )t right ] A_c cos left ( 2 pi f_ct right )$$ $=frac{A_m{A_{c}}^{2}}{2} cosleft [ 2 pileft ( f_c -f_m right )t right ] cosleft ( 2 pi f_ct right )$ $=frac{A_m{A_{c}}^{2}}{4}left { cosleft [ 2 pileft ( 2f_c-fm right ) right ]+ cosleft ( 2 pi f_m right )t right }$ $vleft ( t right )=frac{A_m{A_{c}}^{2}}{4} cosleft ( 2 pi f_mt right )+frac{A_m{A_{c}}^{2}}{4} cosleft [ 2 pi left ( 2f_c-f_m right )t right ]$ In the above equation, the first term is the scaled version of the message signal. It can be extracted by passing the above signal through a low pass filter. Therefore, the output of low pass filter is $$v_0left ( t right )=frac{A_m{A_{c}}^{2}}{4} cosleft ( 2 pi f_mt right )$$ Here, the scaling factor is $frac{{A_{c}}^{2}}{4}$. We can use the same block diagram for demodulating SSBSC wave having an upper sideband. Consider the following SSBSC wave having an upper sideband. $$sleft ( t right )=frac{A_mA_c}{2} cosleft [ 2 pi left ( f_c+f_m right )t right ]$$ The output of the local oscillator is $$cleft ( t right )=A_c cosleft ( 2 pi f_ct right )$$ We can write the output of the product modulator as $$vleft ( t right )=sleft ( t right )cleft ( t right )$$ Substitute $sleft ( t right )$ and $cleft ( t right )$ values in the above equation. $$Rightarrow vleft ( t right )=frac{A_mA_c}{2} cosleft [ 2 pileft ( f_c+f_m right )t right ]A_c cosleft ( 2 pi f_ct right )$$ $=frac{A_m{A_{c}}^{2}}{2} cosleft [ 2 pileft ( f_c+f_m right )t right ] cosleft ( 2 pi f_ct right )$ $=frac{A_m{A_{c}}^{2}}{4} left { cosleft [ 2 pileft ( 2f_c+f_m right )t right ]+ cosleft ( 2 pi f_mt right ) right }$ $vleft ( t right )=frac{A_m{A_{c}}^{2}}{4} cosleft ( 2 pi f_mt right )+frac{A_m{A_{c}}^{2}}{4} cos left [ 2 pileft ( 2f_c+f_m right )t right ]$ In the above equation, the first term is the scaled version of the message signal. It can be extracted by passing the above signal through a low pass filter. Therefore, the output of the low pass filter is $$v_0left ( t right )=frac{A_m{A_{c}}^{2}}{4} cosleft ( 2 pi f_mt right )$$ Here too the scaling factor is $frac{{A_{c}}^{2}}{4}$. Therefore, we get the same demodulated output in both the cases by using coherent detector. Learning working make money

Analog Communication – Noise In any communication system, during the transmission of the signal or while receiving the signal, some unwanted signal gets introduced into the communication, making it unpleasant for the receiver, and questioning the quality of the communication. Such a disturbance is called as Noise. What is Noise? Noise is an unwanted signal, which interferes with the original message signal and corrupts the parameters of the message signal. This alteration in the communication process, leads to the message getting altered. It most likely enters at the channel or the receiver. The noise signal can be understood by taking a look at the following figure. Hence, it is understood that the noise is some signal which has no pattern and no constant frequency or amplitude. It is quite random and unpredictable. Measures are usually taken to reduce it, though it can’t be completely eliminated. Most common examples of noise are − Hiss sound in radio receivers Buzz sound amidst of telephone conversations Flicker in television receivers, etc Types of Noise The classification of noise is done depending on the type of the source, the effect it shows or the relation it has with the receiver, etc. There are two main ways in which noise is produced. One is through some external source while the other is created by an internal source, within the receiver section. External Source This noise is produced by the external sources, which may occur in the medium or channel of communication usually. This noise cannot be completely eliminated. The best way is to avoid the noise from affecting the signal. Examples Most common examples of this type of noise are Atmospheric noise (due to irregularities in the atmosphere). Extra-terrestrial noise, such as solar noise and cosmic noise. Industrial noise. Internal Source This noise is produced by the receiver components while functioning. The components in the circuits, due to continuous functioning, may produce few types of noise. This noise is quantifiable. A proper receiver design may lower the effect of this internal noise. Examples Most common examples of this type of noise are Thermal agitation noise (Johnson noise or Electrical noise) Shot noise (due to the random movement of electrons and holes) Transit-time noise (during transition) Miscellaneous noise is another type of noise which includes flicker, resistance effect and mixer generated noise, etc. Effects of Noise Noise is an inconvenient feature, which affects the system performance. Following are the effects of noise. Noise limits the operating range of the systems Noise indirectly places a limit on the weakest signal that can be amplified by an amplifier. The oscillator in the mixer circuit may limit its frequency because of noise. A system’s operation depends on the operation of its circuits. Noise limits the smallest signal that a receiver is capable of processing. Noise affects the sensitivity of receivers Sensitivity is the minimum amount of input signal necessary to obtain the specified quality output. Noise affects the sensitivity of a receiver system, which eventually affects the output. Learning working make money

Analog Communication – FM Modulators In this chapter, let us discuss about the modulators which generate NBFM and WBFM waves. First, let us discuss about the generation of NBFM. Generation of NBFM We know that the standard equation of FM wave is $$sleft ( t right )=A_c cosleft ( 2 pi f_ct+2 pi k_fint mleft ( t right ) dtright )$$ $Rightarrow sleft ( t right )=A_c cosleft ( 2 pi f_ct right ) cosleft ( 2 pi k_fint mleft ( t right )dt right )-$ $A_c sinleft ( 2 pi f_ct right ) sinleft ( 2 pi k_fint mleft ( t right )dt right )$ For NBFM, $$left | 2 pi k_fint mleft ( t right )dt right | We know that $cos theta approx 1$ and $sin theta approx 1$ when $theta$ is very small. By using the above relations, we will get the NBFM equation as $$sleft ( t right )=A_c cosleft ( 2 pi f_ct right )-A_c sinleft ( 2 pi f_ct right )2 pi k_fint mleft ( t right )dt$$ The block diagram of NBFM modulator is shown in the following figure. Here, the integrator is used to integrate the modulating signal $mleft (t right )$. The carrier signal $A_c cos left ( 2 pi f_ct right )$ is the phase shifted by $-90^0$ to get $A_c sin left ( 2 pi f_ct right )$ with the help of $-90^0$ phase shifter. The product modulator has two inputs $int mleft ( t right )dt$ and $A_c sin left ( 2 pi f_ct right )$. It produces an output, which is the product of these two inputs. This is further multiplied with $2 pi k_f$ by placing a block $2 pi k_f$ in the forward path. The summer block has two inputs, which are nothing but the two terms of NBFM equation. Positive and negative signs are assigned for the carrier signal and the other term at the input of the summer block. Finally, the summer block produces NBFM wave. Generation of WBFM The following two methods generate WBFM wave. Direct method Indirect method Direct Method This method is called as the Direct Method because we are generating a wide band FM wave directly. In this method, Voltage Controlled Oscillator (VCO) is used to generate WBFM. VCO produces an output signal, whose frequency is proportional to the input signal voltage. This is similar to the definition of FM wave. The block diagram of the generation of WBFM wave is shown in the following figure. Here, the modulating signal $mleft (t right )$ is applied as an input of Voltage Controlled Oscillator (VCO). VCO produces an output, which is nothing but the WBFM. $$f_i : alpha : mleft ( t right )$$ $$Rightarrow f_i=f_c+k_fmleft ( t right )$$ Where, $f_i$ is the instantaneous frequency of WBFM wave. Indirect Method This method is called as Indirect Method because we are generating a wide band FM wave indirectly. This means, first we will generate NBFM wave and then with the help of frequency multipliers we will get WBFM wave. The block diagram of generation of WBFM wave is shown in the following figure. This block diagram contains mainly two stages. In the first stage, the NBFM wave will be generated using NBFM modulator. We have seen the block diagram of NBFM modulator at the beginning of this chapter. We know that the modulation index of NBFM wave is less than one. Hence, in order to get the required modulation index (greater than one) of FM wave, choose the frequency multiplier value properly. Frequency multiplier is a non-linear device, which produces an output signal whose frequency is ‘n’ times the input signal frequency. Where, ‘n’ is the multiplication factor. If NBFM wave whose modulation index $beta$ is less than 1 is applied as the input of frequency multiplier, then the frequency multiplier produces an output signal, whose modulation index is ‘n’ times $beta$ and the frequency also ‘n’ times the frequency of WBFM wave. Sometimes, we may require multiple stages of frequency multiplier and mixers in order to increase the frequency deviation and modulation index of FM wave. Learning working make money

Analog Communication – Multiplexing Multiplexing is the process of combining multiple signals into one signal, over a shared medium. If the analog signals are multiplexed, then it is called as analog multiplexing. Similarly, if the digital signals are multiplexed, then it is called as digital multiplexing. Multiplexing was first developed in telephony. A number of signals were combined to send through a single cable. The process of multiplexing divides a communication channel into several number of logical channels, allotting each one for a different message signal or a data stream to be transferred. The device that does multiplexing can be called as Multiplexer or MUX. The reverse process, i.e., extracting the number of channels from one, which is done at the receiver is called as de-multiplexing. The device that does de-multiplexing can be called as de-multiplexer or DEMUX. The following figures illustrates the concept of MUX and DEMUX. Their primary use is in the field of communications. Types of Multiplexers There are mainly two types of multiplexers, namely analog and digital. They are further divided into Frequency Division Multiplexing (FDM), Wavelength Division Multiplexing (WDM), and Time Division Multiplexing (TDM). The following figure gives a detailed idea about this classification. There are many types of multiplexing techniques. Out of which, we have the main types with general classification, mentioned in the above figure. Let us take a look at them individually. Analog Multiplexing The signals used in analog multiplexing techniques are analog in nature. The analog signals are multiplexed according to their frequency (FDM) or wavelength (WDM). Frequency Division Multiplexing In analog multiplexing, the most used technique is Frequency Division Multiplexing (FDM). This technique uses various frequencies to combine streams of data, for sending them on a communication medium, as a single signal. Example − A traditional television transmitter, which sends a number of channels through a single cable uses FDM. Wavelength Division Multiplexing Wavelength Division multiplexing (WDM) is an analog technique, in which many data streams of different wavelengths are transmitted in the light spectrum. If the wavelength increases, the frequency of the signal decreases. A prism, which can turn different wavelengths into a single line, can be used at the output of MUX and input of DEMUX. Example − Optical fiber communications use WDM technique, to merge different wavelengths into a single light for communication. Digital Multiplexing The term digital represents the discrete bits of information. Hence, the available data is in the form of frames or packets, which are discrete. Time Division Multiplexing In Time Division Multiplexing (TDM), the time frame is divided into slots. This technique is used to transmit a signal over a single communication channel, by allotting one slot for each message. Time Division Multiplexing (TDM) can be classified into Synchronous TDM and Asynchronous TDM. Synchronous TDM In Synchronous TDM, the input is connected to a frame. If there are ‘n’ number of connections, then the frame is divided into ‘n’ time slots. One slot is allocated for each input line. In this technique, the sampling rate is common for all signals and hence the same clock input is given. The MUX allocates the same slot to each device at all times. Asynchronous TDM In Asynchronous TDM, the sampling rate is different for each of the signals and a common clock is not required. If the allotted device for a time slot transmits nothing and sits idle, then that slot can be allotted to another device, unlike synchronous This type of TDM is used in Asynchronous transfer mode networks. De-Multiplexer De-multiplexers are used to connect a single source to multiple destinations. This process is the reverse process of multiplexing. As mentioned previously, it is used mostly at the receivers. DEMUX has many applications. It is used in receivers in the communication systems. It is used in arithmetic and logical unit in computers to supply power and to pass on communication, etc. De-multiplexers are used as serial to parallel converters. The serial data is given as input to DEMUX at regular interval and a counter is attached to it to control the output of the de-multiplexer. Both the multiplexers and de-multiplexers play an important role in communication systems, both at the transmitter and the receiver sections. Learning working make money

Analog Communication – Angle Modulation The other type of modulation in continuous-wave modulation is Angle Modulation. Angle Modulation is the process in which the frequency or the phase of the carrier signal varies according to the message signal. The standard equation of the angle modulated wave is $$sleft ( t right )=A_c cos theta _ileft ( t right )$$ Where, $A_c$ is the amplitude of the modulated wave, which is the same as the amplitude of the carrier signal $theta _ileft ( t right )$ is the angle of the modulated wave Angle modulation is further divided into frequency modulation and phase modulation. Frequency Modulation is the process of varying the frequency of the carrier signal linearly with the message signal. Phase Modulation is the process of varying the phase of the carrier signal linearly with the message signal. Now, let us discuss these in detail. Frequency Modulation In amplitude modulation, the amplitude of the carrier signal varies. Whereas, in Frequency Modulation (FM), the frequency of the carrier signal varies in accordance with the instantaneous amplitude of the modulating signal. Hence, in frequency modulation, the amplitude and the phase of the carrier signal remains constant. This can be better understood by observing the following figures. The frequency of the modulated wave increases, when the amplitude of the modulating or message signal increases. Similarly, the frequency of the modulated wave decreases, when the amplitude of the modulating signal decreases. Note that, the frequency of the modulated wave remains constant and it is equal to the frequency of the carrier signal, when the amplitude of the modulating signal is zero. Mathematical Representation The equation for instantaneous frequency $f_i$ in FM modulation is $$f_i=f_c+k_fmleft ( t right )$$ Where, $f_c$ is the carrier frequency $k_t$ is the frequency sensitivity $mleft ( t right )$ is the message signal We know the relationship between angular frequency $omega_i$ and angle $theta _ileft ( t right )$ as $$omega_i=frac{dtheta _ileft ( t right )}{dt}$$ $Rightarrow 2 pi f_i=frac{dtheta _ileft ( t right )}{dt}$ $Rightarrow theta _ileft ( t right )= 2piint f_i dt$ Substitute, $f_i$ value in the above equation. $$theta _ileft ( t right )=2 piint left ( f_c+k_f mleft ( t right ) right )dt$$ $Rightarrow theta _ileft ( t right )=2 pi f_ct+2 pi k_fint mleft ( t right )dt$ Substitute, $theta _ileft ( t right )$ value in the standard equation of angle modulated wave. $$sleft ( t right )=A_c cosleft ( 2 pi f_ct + 2 pi k_f int mleft ( t right )dt right )$$ This is the equation of FM wave. If the modulating signal is $mleft ( t right )= A_m cos left ( 2 pi f_mt right )$, then the equation of FM wave will be $$sleft ( t right )=A_c cosleft ( 2 pi f_ct + beta sin left ( 2 pi f_mt right ) right )$$ Where, $beta$ = modulation index $=frac{Delta f}{f_m}=frac{k_fA_m}{f_m}$ The difference between FM modulated frequency (instantaneous frequency) and normal carrier frequency is termed as Frequency Deviation. It is denoted by $Delta f$, which is equal to the product of $k_f$ and $A_m$. FM can be divided into Narrowband FM and Wideband FM based on the values of modulation index $beta$. Narrowband FM Following are the features of Narrowband FM. This frequency modulation has a small bandwidth when compared to wideband FM. The modulation index $beta$ is small, i.e., less than 1. Its spectrum consists of the carrier, the upper sideband and the lower sideband. This is used in mobile communications such as police wireless, ambulances, taxicabs, etc. Wideband FM Following are the features of Wideband FM. This frequency modulation has infinite bandwidth. The modulation index $beta$ is large, i.e., higher than 1. Its spectrum consists of a carrier and infinite number of sidebands, which are located around it. This is used in entertainment, broadcasting applications such as FM radio, TV, etc. Phase Modulation In frequency modulation, the frequency of the carrier varies. Whereas, in Phase Modulation (PM), the phase of the carrier signal varies in accordance with the instantaneous amplitude of the modulating signal. So, in phase modulation, the amplitude and the frequency of the carrier signal remains constant. This can be better understood by observing the following figures. The phase of the modulated wave has got infinite points, where the phase shift in a wave can take place. The instantaneous amplitude of the modulating signal changes the phase of the carrier signal. When the amplitude is positive, the phase changes in one direction and if the amplitude is negative, the phase changes in the opposite direction. Mathematical Representation The equation for instantaneous phase $phi_i$ in phase modulation is $$phi _i=k_p mleft ( t right )$$ Where, $k_p$ is the phase sensitivity $mleft ( t right )$ is the message signal The standard equation of angle modulated wave is $$sleft ( t right )=A_c cos left ( 2 pi f_ct+phi_i right )$$ Substitute, $phi_i$ value in the above equation. $$sleft ( t right )=A_c cos left ( 2 pi f_ct+k_p m left ( t right )right )$$ This is the equation of PM wave. If the modulating signal, $mleft ( t right )=A_m cos left ( 2 pi f_mt right ) $, then the equation of PM wave will be $$sleft ( t right )=A_c cosleft (2 pi f_ct+beta cosleft ( 2 pi f_mt right ) right )$$ Where, $beta$ = modulation index = $Delta phi=k_pA_m$ $Delta phi$ is phase deviation Phase modulation is used in mobile communication systems, while frequency modulation is used mainly for FM broadcasting. Learning working make money

Analog Communication – AM Demodulators The process of extracting an original message signal from the modulated wave is known as detection or demodulation. The circuit, which demodulates the modulated wave is known as the demodulator. The following demodulators (detectors) are used for demodulating AM wave. Square Law Demodulator Envelope Detector Square Law Demodulator Square law demodulator is used to demodulate low level AM wave. Following is the block diagram of thesquare law demodulator. This demodulator contains a square law device and low pass filter. The AM wave $V_1left ( t right )$ is applied as an input to this demodulator. The standard form of AM wave is $$V_1left ( t right )=A_cleft [ 1+k_amleft ( t right ) right ] cosleft ( 2 pi f_ct right )$$ We know that the mathematical relationship between the input and the output of square law device is $V_2left ( t right )=k_1V_1left ( t right )+k_2V_1^2left ( t right )$(Equation 1) Where, $V_1left ( t right )$ is the input of the square law device, which is nothing but the AM wave $V_2left ( t right )$ is the output of the square law device $k_1$ and $k_2$ are constants Substitute $V_1left ( t right )$ in Equation 1 $$V_2left ( t right )=k_1left ( A_cleft [ 1+k_amleft ( t right ) right ] cosleft ( 2 pi f_ct right )right )+k_2left ( A_cleft [ 1+k_amleft ( t right ) right ] cosleft ( 2 pi f_ct right )right )^2$$ $Rightarrow V_2left ( t right )=k_1A_c cos left ( 2 pi f_ct right )+k_1A_ck_amleft ( t right ) cos left ( 2 pi f_ct right )+$ $k_2{A_{c}}^{2}left [ 1+{K_{a}}^{2}m^2left ( t right )+2k_amleft ( t right ) right ]left ( frac{1+ cosleft ( 4 pi f_ct right )}{2} right )$ $Rightarrow V_2left ( t right )=k_1A_c cos left ( 2 pi f_ct right )+k_1A_ck_amleft ( t right ) cos left ( 2 pi f_ct right)+frac{K_2{A_{c}}^{2}}{2}+$ $frac{K_2{A_{c}}^{2}}{2} cos left ( 4 pi f_ct right )+frac{k_2 {A_{c}}^{2}{k_{a}}^{2}m^2left ( t right )}{2}+frac{k_2 {A_{c}}^{2}{k_{a}}^{2}m^2left ( t right )}{2} cosleft ( 4 pi f_ct right )+$ $k_2{A_{c}}^{2}k_amleft ( t right )+k_2{A_{c}}^{2}k_amleft ( t right )cos left ( 4 pi f_ct right )$ In the above equation, the term $k_2{A_{c}}^{2}k_amleft ( t right )$ is the scaled version of the message signal. It can be extracted by passing the above signal through a low pass filter and the DC component $frac{k_2{A_{c}}^{2}}{2}$ can be eliminated with the help of a coupling capacitor. Envelope Detector Envelope detector is used to detect (demodulate) high level AM wave. Following is the block diagram of the envelope detector. This envelope detector consists of a diode and low pass filter. Here, the diode is the main detecting element. Hence, the envelope detector is also called as the diode detector. The low pass filter contains a parallel combination of the resistor and the capacitor. The AM wave $sleft ( t right )$ is applied as an input to this detector. We know the standard form of AM wave is $$sleft ( t right )=A_cleft [ 1+k_amleft ( t right ) right ] cosleft ( 2 pi f_ct right )$$ In the positive half cycle of AM wave, the diode conducts and the capacitor charges to the peak value of AM wave. When the value of AM wave is less than this value, the diode will be reverse biased. Thus, the capacitor will discharge through resistor R till the next positive half cycle of AM wave. When the value of AM wave is greater than the capacitor voltage, the diode conducts and the process will be repeated. We should select the component values in such a way that the capacitor charges very quickly and discharges very slowly. As a result, we will get the capacitor voltage waveform same as that of the envelope of AM wave, which is almost similar to the modulating signal. Learning working make money

Analog Communication – SSBSC Modulators In this chapter, let us discuss about the modulators, which generate SSBSC wave. We can generate SSBSC wave using the following two methods. Frequency discrimination method Phase discrimination method Frequency Discrimination Method The following figure shows the block diagram of SSBSC modulator using frequency discrimination method. In this method, first we will generate DSBSC wave with the help of the product modulator. Then, apply this DSBSC wave as an input of band pass filter. This band pass filter produces an output, which is SSBSC wave. Select the frequency range of band pass filter as the spectrum of the desired SSBSC wave. This means the band pass filter can be tuned to either upper sideband or lower sideband frequencies to get the respective SSBSC wave having upper sideband or lower sideband. Phase Discrimination Method The following figure shows the block diagram of SSBSC modulator using phase discrimination method. This block diagram consists of two product modulators, two $-90^0$ phase shifters, one local oscillator and one summer block. The product modulator produces an output, which is the product of two inputs. The $-90^0$ phase shifter produces an output, which has a phase lag of $-90^0$ with respect to the input. The local oscillator is used to generate the carrier signal. Summer block produces an output, which is either the sum of two inputs or the difference of two inputs based on the polarity of inputs. The modulating signal $A_m cosleft ( 2 pi f_mt right )$ and the carrier signal $A_c cosleft ( 2 pi f_ct right )$ are directly applied as inputs to the upper product modulator. So, the upper product modulator produces an output, which is the product of these two inputs. The output of upper product modulator is $$s_1left ( t right )=A_mA_c cos left ( 2 pi f_mt right ) cosleft ( 2 pi f_ct right )$$ $$ Rightarrow s_1left ( t right )=frac{A_mA_c}{2} left { cos left [ 2 pileft ( f_c+f_m right )t right ]+ cosleft [ 2 pileft ( f_c-f_m right )t right ] right }$$ The modulating signal $A_m cosleft ( 2 pi f_mt right )$ and the carrier signal $A_c cosleft ( 2 pi f_ct right )$ are phase shifted by $-90^0$ before applying as inputs to the lower product modulator. So, the lower product modulator produces an output, which is the product of these two inputs. The output of lower product modulator is $$s_2left ( t right )=A_mA_c cosleft ( 2 pi f_mt-90^0 right ) cosleft (2 pi f_ct-90^0 right )$$ $Rightarrow s_2left ( t right )=A_mA_c sin left ( 2 pi f_mt right )sin left ( 2 pi f_ct right )$ $Rightarrow s_2left ( t right )=frac{A_mA_c}{2} left { cos left [ 2 pileft ( f_c-f_m right )t right ]- cosleft [ 2 pileft ( f_c+f_m right )t right ] right }$ Add $s_1left ( t right )$ and $s_2left ( t right )$ in order to get the SSBSC modulated wave $sleft ( t right )$ having a lower sideband. $sleft ( t right )=frac{A_mA_c}{2}left { cosleft [ 2 pileft ( f_c+f_m right )t right ]+cosleft [ 2 pileft ( f_c-f_m right )t right ] right }+$ $frac{A_mA_c}{2}left { cosleft [ 2 pileft ( f_c-f_m right )t right ]-cosleft [ 2 pileft ( f_c+f_m right )t right ] right }$ $Rightarrow sleft ( t right )=A_mA_c cos left [ 2 pileft ( f_c-f_m right )t right ]$ Subtract $s_2left ( t right )$ from $s_1left ( t right )$ in order to get the SSBSC modulated wave $sleft ( t right )$ having a upper sideband. $sleft ( t right )=frac{A_mA_c}{2}left { cosleft [ 2 pileft ( f_c+f_m right )t right ]+cosleft [ 2 pileft ( f_c-f_m right )t right ] right }-$ $frac{A_mA_c}{2}left { cosleft [ 2 pileft ( f_c-f_m right )t right ]-cosleft [ 2 pileft ( f_c+f_m right )t right ] right }$ $Rightarrow sleft ( t right )=A_mA_c cos left [ 2 pileft ( f_c+f_m right )t right ]$ Hence, by properly choosing the polarities of inputs at summer block, we will get SSBSC wave having a upper sideband or a lower sideband. Learning working make money

Numerical Problems 1 In the previous chapter, we have discussed the parameters used in Amplitude Modulation. Each parameter has its own formula. By using those formulas, we can find the respective parameter values. In this chapter, let us solve a few problems based on the concept of amplitude modulation. Problem 1 A modulating signal $mleft ( t right )=10 cos left ( 2pi times 10^3 tright )$ is amplitude modulated with a carrier signal $cleft ( t right )=50 cos left ( 2pi times 10^5 tright )$. Find the modulation index, the carrier power, and the power required for transmitting AM wave. Solution Given, the equation of modulating signal as $$mleft ( t right )=10cos left ( 2pi times 10^3 tright )$$ We know the standard equation of modulating signal as $$mleft ( t right )=A_mcosleft ( 2pi f_mt right )$$ By comparing the above two equations, we will get Amplitude of modulating signal as $A_m=10 volts$ and Frequency of modulating signal as $$f_m=10^3 Hz=1 KHz$$ Given, the equation of carrier signal is $$cleft ( t right )=50cos left ( 2pi times 10^5t right )$$ The standard equation of carrier signal is $$cleft ( t right )=A_ccosleft ( 2pi f_ct right )$$ By comparing these two equations, we will get Amplitude of carrier signal as $A_c=50volts$ and Frequency of carrier signal as $f_c=10^5 Hz=100 KHz$ We know the formula for modulation index as $$mu =frac{A_m}{A_c}$$ Substitute, $A_m$ and $A_c$ values in the above formula. $$mu=frac{10}{50}=0.2$$ Therefore, the value of modulation index is 0.2 and percentage of modulation is 20%. The formula for Carrier power, $P_c=$ is $$P_c=frac{{A_{c}}^{2}}{2R}$$ Assume $R=1Omega$ and substitute $A_c$ value in the above formula. $$P_c=frac{left ( 50 right )^2}{2left ( 1 right )}=1250W$$ Therefore, the Carrier power, $P_c$ is 1250 watts. We know the formula for power required for transmitting AM wave is $$Rightarrow P_t=P_cleft ( 1+frac{mu ^2}{2} right )$$ Substitute $P_c$ and $mu$ values in the above formula. $$P_t=1250left ( 1+frac{left ( 0.2 right )^2}{2} right )=1275W$$ Therefore, the power required for transmitting AM wave is 1275 watts. Problem 2 The equation of amplitude wave is given by $sleft ( t right ) = 20left [ 1 + 0.8 cos left ( 2pi times 10^3t right ) right ]cos left ( 4pi times 10^5t right )$. Find the carrier power, the total sideband power, and the band width of AM wave. Solution Given, the equation of Amplitude modulated wave is $$sleft ( t right )=20left [ 1+0.8 cosleft ( 2pi times 10^3t right ) right ]cos left ( 4pi times 10^5t right )$$ Re-write the above equation as $$sleft ( t right )=20left [ 1+0.8 cosleft ( 2pi times 10^3t right ) right ]cos left ( 2pi times 2 times 10^5t right )$$ We know the equation of Amplitude modulated wave is $$sleft ( t right )=A_cleft [ 1+mu cosleft ( 2pi f_mt right ) right ]cosleft ( 2 pi f_ct right )$$ By comparing the above two equations, we will get Amplitude of carrier signal as $A_c=20 volts$ Modulation index as $mu=0.8$ Frequency of modulating signal as $f_m=10^3Hz=1 KHz$ Frequency of carrier signal as $f_c=2times 10^5Hz=200KHz$ The formula for Carrier power, $P_c$is $$P_c=frac{{A_{e}}^{2}}{2R}$$ Assume $R=1Omega$ and substitute $A_c$ value in the above formula. $$P_c=frac{left ( 20 right )^2}{2left ( 1 right )}=200W$$ Therefore, the Carrier power, $P_c$ is 200watts. We know the formula for total side band power is $$P_{SB}=frac{P_cmu^2}{2}$$ Substitute $P_c$ and $mu$ values in the above formula. $$P_{SB}=frac{200times left ( 0.8 right )^2}{2}=64W$$ Therefore, the total side band power is 64 watts. We know the formula for bandwidth of AM wave is $$BW=2f_m$$ Substitute $f_m$ value in the above formula. $$BW=2left ( 1K right )=2 KHz$$ Therefore, the bandwidth of AM wave is 2 KHz. Learning working make money

Analog Communication – DSBSC Modulation In the process of Amplitude Modulation, the modulated wave consists of the carrier wave and two sidebands. The modulated wave has the information only in the sidebands. Sideband is nothing but a band of frequencies, containing power, which are the lower and higher frequencies of the carrier frequency. The transmission of a signal, which contains a carrier along with two sidebands can be termed as Double Sideband Full Carrier system or simply DSBFC. It is plotted as shown in the following figure. However, such a transmission is inefficient. Because, two-thirds of the power is being wasted in the carrier, which carries no information. If this carrier is suppressed and the saved power is distributed to the two sidebands, then such a process is called as Double Sideband Suppressed Carrier system or simply DSBSC. It is plotted as shown in the following figure. Mathematical Expressions Let us consider the same mathematical expressions for modulating and carrier signals as we have considered in the earlier chapters. i.e., Modulating signal $$mleft ( t right )=A_m cos left ( 2 pi f_mtright )$$ Carrier signal $$cleft ( t right )=A_c cos left ( 2 pi f_ctright )$$ Mathematically, we can represent the equation of DSBSC wave as the product of modulating and carrier signals. $$sleft ( t right )=mleft ( t right )cleft ( t right )$$ $$Rightarrow sleft ( t right )=A_mA_c cos left ( 2 pi f_mt right )cos left ( 2 pi f_ct right )$$ Bandwidth of DSBSC Wave We know the formula for bandwidth (BW) is $$BW=f_{max}-f_{min}$$ Consider the equation of DSBSC modulated wave. $$sleft ( t right )=A_mA_c cosleft ( 2 pi f_mt right ) cos(2 pi f_ct)$$ $$Rightarrow sleft ( t right )=frac{A_mA_c}{2} cosleft [ 2 pileft ( f_c+f_m right ) tright ]+frac{A_mA_c}{2} cosleft [ 2 pileft ( f_c-f_m right ) tright ]$$ The DSBSC modulated wave has only two frequencies. So, the maximum and minimum frequencies are $f_c+f_m$ and $f_c-f_m$ respectively. i.e., $f_{max}=f_c+f_m$ and $f_{min}=f_c-f_m$ Substitute, $f_{max}$ and $f_{min}$ values in the bandwidth formula. $$BW=f_c+f_m-left ( f_c-f_m right )$$ $$Rightarrow BW=2f_m$$ Thus, the bandwidth of DSBSC wave is same as that of AM wave and it is equal to twice the frequency of the modulating signal. Power Calculations of DSBSC Wave Consider the following equation of DSBSC modulated wave. $$sleft ( t right )=frac{A_mA_c}{2} cosleft [ 2 pi left ( f_c+f_m right ) tright ]+frac{A_mA_c}{2} cosleft [ 2 pi left ( f_c-f_m right ) tright ]$$ Power of DSBSC wave is equal to the sum of powers of upper sideband and lower sideband frequency components. $$P_t=P_{USB}+P_{LSB}$$ We know the standard formula for power of cos signal is $$P=frac{{v_{rms}}^{2}}{R}=frac{left ( v_msqrt{2}right )^2}{R}$$ First, let us find the powers of upper sideband and lower sideband one by one. Upper sideband power $$P_{USB}=frac{left ( A_mA_c / 2sqrt{2}right )^2}{R}=frac{{A_{m}}^{2}{A_{c}}^{2}}{8R}$$ Similarly, we will get the lower sideband power same as that of upper sideband power. $$P_{USB}=frac{{A_{m}}^{2}{A_{c}}^{2}}{8R}$$ Now, let us add these two sideband powers in order to get the power of DSBSC wave. $$P_t=frac{{A_{m}}^{2}{A_{c}}^{2}}{8R}+frac{{A_{m}}^{2}{A_{c}}^{2}}{8R}$$ $$Rightarrow P_t=frac{{A_{m}}^{2}{A_{c}}^{2}}{4R}$$ Therefore, the power required for transmitting DSBSC wave is equal to the power of both the sidebands. Learning working make money