Category: digital Signal Processing

DSP – Classification of DT Signals Just like Continuous time signals, Discrete time signals can be classified according to the conditions or operations on the signals. Even and Odd Signals Even Signal A signal is said to be even or symmetric if it satisfies the following condition; $$x(-n) = x(n)$$ Here, we can see that x(-1) = x(1), x(-2) = x(2) and x(-n) = x(n). Thus, it is an even signal. Odd Signal A signal is said to be odd if it satisfies the following condition; $$x(-n) = -x(n)$$ From the figure, we can see that x(1) = -x(-1), x(2) = -x(2) and x(n) = -x(-n). Hence, it is an odd as well as anti-symmetric signal. Periodic and Non-Periodic Signals A discrete time signal is periodic if and only if, it satisfies the following condition − $$x(n+N) = x(n)$$ Here, x(n) signal repeats itself after N period. This can be best understood by considering a cosine signal − $$x(n) = A cos(2pi f_{0}n+theta)$$ $$x(n+N) = Acos(2pi f_{0}(n+N)+theta) = Acos(2pi f_{0}n+2pi f_{0}N+theta)$$ $$= Acos(2pi f_{0}n+2pi f_{0}N+theta)$$ For the signal to become periodic, following condition should be satisfied; $$x(n+N) = x(n)$$ $$Rightarrow Acos(2pi f_{0}n+2pi f_{0}N+theta) = A cos(2pi f_{0}n+theta)$$ i.e. $2pi f_{0}N$ is an integral multiple of $2pi$ $$2pi f_{0}N = 2pi K$$ $$Rightarrow N = frac{K}{f_{0}}$$ Frequencies of discrete sinusoidal signals are separated by integral multiple of $2pi$. Energy and Power Signals Energy Signal Energy of a discrete time signal is denoted as E. Mathematically, it can be written as; $$E = displaystyle sumlimits_{n=-infty}^{+infty}|x(n)|^2$$ If each individual values of $x(n)$ are squared and added, we get the energy signal. Here $x(n)$ is the energy signal and its energy is finite over time i.e $0 Power Signal Average power of a discrete signal is represented as P. Mathematically, this can be written as; $$P = lim_{N to infty} frac{1}{2N+1}displaystylesumlimits_{n=-N}^{+N} |x(n)|^2$$ Here, power is finite i.e. 0<P<∞. However, there are some signals, which belong to neither energy nor power type signal. Learning working make money

DSP – Miscellaneous Signals There are other signals, which are a result of operation performed on them. Some common type of signals are discussed below. Conjugate Signals Signals, which satisfies the condition $x(t) = x*(-t)$ are called conjugate signals. Let $x(t) = a(t)+jb(t)$…eqn. 1 So, $x(-t) = a(-t)+jb(-t)$ And $x*(-t) = a(-t)-jb(-t)$…eqn. 2 By Condition, $x(t) = x*(-t)$ If we compare both the derived equations 1 and 2, we can see that the real part is even, whereas the imaginary part is odd. This is the condition for a signal to be a conjugate type. Conjugate Anti-Symmetric Signals Signals, which satisfy the condition $x(t) = -x*(-t)$ are called conjugate anti-symmetric signal Let $x(t) = a(t)+jb(t)$…eqn. 1 So $x(-t) = a(-t)+jb(-t)$ And $x*(-t) = a(-t)-jb(-t)$ $-x*(-t) = -a(-t)+jb(-t)$…eqn. 2 By Condition $x(t) = -x*(-t)$ Now, again compare, both the equations just as we did for conjugate signals. Here, we will find that the real part is odd and the imaginary part is even. This is the condition for a signal to become conjugate anti-symmetric type. Example Let the signal given be $x(t) = sin t+jt^{2}$. Here, the real part being $sin t$ is odd and the imaginary part being $t^2$ is even. So, this signal can be classified as conjugate anti-symmetric signal. Any function can be divided into two parts. One part being Conjugate symmetry and other part being conjugate anti-symmetric. So any signal x(t) can be written as $$x(t) = xcs(t)+xcas(t)$$ Where $xcs(t)$ is conjugate symmetric signal and $xcas(t)$ is conjugate anti symmetric signal $$xcs(t) = frac{[x(t)+x*(-t)]}{2}$$ And $$xcas(t) = frac{[x(t)-x*(-t)]}{2}$$ Half Wave Symmetric Signals When a signal satisfies the condition $cx(t) = -x(tpm (frac{T_{0}}{2}))$, it is called half wave symmetric signal. Here, amplitude reversal and time shifting of the signal takes place by half time. For half wave symmetric signal, average value will be zero but this is not the case when the situation is reversed. Consider a signal x(t) as shown in figure A above. The first step is to time shift the signal and make it $x[t-(frac{T}{2})]$. So, the new signal is changed as shown in figure B. Next, we reverse the amplitude of the signal, i.e. make it $-x[t-(frac{T}{2})]$ as shown in figure C. Since, this signal repeats itself after half-time shifting and reversal of amplitude, it is a half wave symmetric signal. Orthogonal Signal Two signals x(t) and y(t) are said to be orthogonal if they satisfy the following two conditions. Condition 1 − $int_{-infty}^{infty}x(t)y(t) = 0$ [for non-periodic signal] Condition 2 − $int x(t)y(t) = 0$ [For periodic Signal] The signals, which contain odd harmonics (3rd, 5th, 7th …etc.) and have different frequencies, are mutually orthogonal to each other. In trigonometric type signals, sine functions and cosine functions are also orthogonal to each other; provided, they have same frequency and are in same phase. In the same manner DC (Direct current signals) and sinusoidal signals are also orthogonal to each other. If x(t) and y(t) are two orthogonal signals and $z(t) = x(t)+y(t)$ then the power and energy of z(t) can be written as ; $$P(z) = p(x)+p(y)$$ $$E(z) = E(x)+E(y)$$ Example Analyze the signal: $z(t) = 3+4sin(2pi t+30^0)$ Here, the signal comprises of a DC signal (3) and one sine function. So, by property this signal is an orthogonal signal and the two sub-signals in it are mutually orthogonal to each other. Learning working make money

Digital Signal Processing – Basic DT Signals We have seen that how the basic signals can be represented in Continuous time domain. Let us see how the basic signals can be represented in Discrete Time Domain. Unit Impulse Sequence It is denoted as δ(n) in discrete time domain and can be defined as; $$delta(n)=begin{cases}1, & for quad n=0\0, & Otherwiseend{cases}$$ Unit Step Signal Discrete time unit step signal is defined as; $$U(n)=begin{cases}1, & for quad ngeq0\0, & for quad n<0end{cases}$$ The figure above shows the graphical representation of a discrete step function. Unit Ramp Function A discrete unit ramp function can be defined as − $$r(n)=begin{cases}n, & for quad ngeq0\0, & for quad n<0end{cases}$$ The figure given above shows the graphical representation of a discrete ramp signal. Parabolic Function Discrete unit parabolic function is denoted as p(n) and can be defined as; $$p(n) = begin{cases}frac{n^{2}}{2} ,& for quad ngeq0\0, & for quad n<0end{cases}$$ In terms of unit step function it can be written as; $$P(n) = frac{n^{2}}{2}U(n)$$ The figure given above shows the graphical representation of a parabolic sequence. Sinusoidal Signal All continuous-time signals are periodic. The discrete-time sinusoidal sequences may or may not be periodic. They depend on the value of ω. For a discrete time signal to be periodic, the angular frequency ω must be a rational multiple of 2π. A discrete sinusoidal signal is shown in the figure above. Discrete form of a sinusoidal signal can be represented in the format − $$x(n) = Asin(omega n + phi)$$ Here A,ω and φ have their usual meaning and n is the integer. Time period of the discrete sinusoidal signal is given by − $$N =frac{2pi m}{omega}$$ Where, N and m are integers. Learning working make money

Digital Signal Processing – Quick Guide Digital Signal Processing – Signals-Definition Definition Anything that carries information can be called as signal. It can also be defined as a physical quantity that varies with time, temperature, pressure or with any independent variables such as speech signal or video signal. The process of operation in which the characteristics of a signal (Amplitude, shape, phase, frequency, etc.) undergoes a change is known as signal processing. Note − Any unwanted signal interfering with the main signal is termed as noise. So, noise is also a signal but unwanted. According to their representation and processing, signals can be classified into various categories details of which are discussed below. Continuous Time Signals Continuous-time signals are defined along a continuum of time and are thus, represented by a continuous independent variable. Continuous-time signals are often referred to as analog signals. This type of signal shows continuity both in amplitude and time. These will have values at each instant of time. Sine and cosine functions are the best example of Continuous time signal. The signal shown above is an example of continuous time signal because we can get value of signal at each instant of time. Discrete Time signals The signals, which are defined at discrete times are known as discrete signals. Therefore, every independent variable has distinct value. Thus, they are represented as sequence of numbers. Although speech and video signals have the privilege to be represented in both continuous and discrete time format; under certain circumstances, they are identical. Amplitudes also show discrete characteristics. Perfect example of this is a digital signal; whose amplitude and time both are discrete. The figure above depicts a discrete signal’s discrete amplitude characteristic over a period of time. Mathematically, these types of signals can be formularized as; $$x = left { xleft [ n right ] right },quad -infty < n< infty$$ Where, n is an integer. It is a sequence of numbers x, where nth number in the sequence is represented as x[n]. Digital Signal Processing – Basic CT Signals To test a system, generally, standard or basic signals are used. These signals are the basic building blocks for many complex signals. Hence, they play a very important role in the study of signals and systems. Unit Impulse or Delta Function A signal, which satisfies the condition, $delta(t) = lim_{epsilon to infty} x(t)$ is known as unit impulse signal. This signal tends to infinity when t = 0 and tends to zero when t ≠ 0 such that the area under its curve is always equals to one. The delta function has zero amplitude everywhere excunit_impulse.jpgept at t = 0. Properties of Unit Impulse Signal δ(t) is an even signal. δ(t) is an example of neither energy nor power (NENP) signal. Area of unit impulse signal can be written as; $$A = int_{-infty}^{infty} delta (t)dt = int_{-infty}^{infty} lim_{epsilon to 0} x(t) dt = lim_{epsilon to 0} int_{-infty}^{infty} [x(t)dt] = 1$$ Weight or strength of the signal can be written as; $$y(t) = Adelta (t)$$ Area of the weighted impulse signal can be written as − $$y (t) = int_{-infty}^{infty} y (t)dt = int_{-infty}^{infty} Adelta (t) = A[int_{-infty}^{infty} delta (t)dt ] = A = 1 = Wigthedimpulse$$ Unit Step Signal A signal, which satisfies the following two conditions − $U(t) = 1(whenquad t geq 0 )and$ $U(t) = 0 (whenquad t < 0 )$ is known as a unit step signal. It has the property of showing discontinuity at t = 0. At the point of discontinuity, the signal value is given by the average of signal value. This signal has been taken just before and after the point of discontinuity (according to Gibb’s Phenomena). If we add a step signal to another step signal that is time scaled, then the result will be unity. It is a power type signal and the value of power is 0.5. The RMS (Root mean square) value is 0.707 and its average value is also 0.5 Ramp Signal Integration of step signal results in a Ramp signal. It is represented by r(t). Ramp signal also satisfies the condition $r(t) = int_{-infty}^{t} U(t)dt = tU(t)$. It is neither energy nor power (NENP) type signal. Parabolic Signal Integration of Ramp signal leads to parabolic signal. It is represented by p(t). Parabolic signal also satisfies he condition $p(t) = int_{-infty}^{t} r(t)dt = (t^{2}/2)U(t)$ . It is neither energy nor Power (NENP) type signal. Signum Function This function is represented as $$sgn(t) = begin{cases}1 & forquad t >0\-1 & forquad t<0end{cases}$$ It is a power type signal. Its power value and RMS (Root mean square) values, both are 1. Average value of signum function is zero. Sinc Function It is also a function of sine and is written as − $$SinC(t) = frac{SinPi t}{Pi T} = Sa(Pi t)$$ Properties of Sinc function It is an energy type signal. $Sinc(0) = lim_{t to 0}frac{sin Pi t}{Pi t} = 1$ $Sinc(infty) = lim_{t to infty}frac{sin Pi infty}{Pi infty} = 0$ (Range of sinπ∞ varies between -1 to +1 but anything divided by infinity is equal to zero) If $ sin c(t) = 0 => sin Pi t = 0$ $Rightarrow Pi t = nPi$ $Rightarrow t = n (n neq 0)$ Sinusoidal Signal A signal, which is continuous in nature is known as continuous signal. General format of a sinusoidal signal is $$x(t) = Asin (omega t + phi )$$ Here, A = amplitude of the signal ω = Angular frequency of the signal (Measured in radians) φ = Phase angle of the signal (Measured in radians) The tendency of this signal is to repeat itself after certain period of time, thus is called periodic signal. The time period of signal is given as; $$T = frac{2pi }{omega }$$ The diagrammatic view of sinusoidal signal is shown below. Rectangular Function A signal is said to be rectangular function type if it satisfies the following condition − $$pi(frac{t}{tau}) = begin{cases}1, & forquad tleq frac{tau}{2}\0, & Otherwiseend{cases}$$ Being symmetrical about

DSP – Fast Fourier Transform In earlier DFT methods, we have seen that the computational part is too long. We want to reduce that. This can be done through FFT or fast Fourier transform. So, we can say FFT is nothing but computation of discrete Fourier transform in an algorithmic format, where the computational part will be reduced. The main advantage of having FFT is that through it, we can design the FIR filters. Mathematically, the FFT can be written as follows; $$x[K] = displaystylesumlimits_{n = 0}^{N-1}x[n]W_N^{nk}$$ Let us take an example to understand it better. We have considered eight points named from $x_0quad toquad x_7$. We will choose the even terms in one group and the odd terms in the other. Diagrammatic view of the above said has been shown below − Here, points x0, x2, x4 and x6 have been grouped into one category and similarly, points x1, x3, x5 and x7 has been put into another category. Now, we can further make them in a group of two and can proceed with the computation. Now, let us see how these breaking into further two is helping in computation. $x[k] = displaystylesumlimits_{r = 0}^{frac{N}{2}-1}x[2r]W_N^{2rk}+displaystylesumlimits_{r = 0}^{frac{N}{2}-1}x[2r+1]W_N^{(2r+1)k}$ $= sum_{r = 0}^{frac{N}{2}-1}x[2r]W_{N/2}^{rk}+sum_{r = 0}^{frac{N}{2}-1}x[2r+1]W_{N/2}^{rk}times W_N^k$ $= G[k]+H[k]times W_N^k$ Initially, we took an eight-point sequence, but later we broke that one into two parts G[k] and H[k]. G[k] stands for the even part whereas H[k] stands for the odd part. If we want to realize it through a diagram, then it can be shown as below − From the above figure, we can see that $W_8^4 = -1$ $W_8^5 = -W_8^1$ $W_8^6 = -W_8^2$ $W_8^7 = -W_8^3$ Similarly, the final values can be written as follows − $G[0]-H[0] = x[4]$ $G[1]-W_8^1H[1] = x[5]$ $G[2]-W_8^2H[2] = x[6]$ $G[1]-W_8^3H[3] = x[7]$ The above one is a periodic series. The disadvantage of this system is that K cannot be broken beyond 4 point. Now Let us break down the above into further. We will get the structures something like this Example Consider the sequence x[n]={ 2,1,-1,-3,0,1,2,1}. Calculate the FFT. Solution − The given sequence is x[n]={ 2,1,-1,-3,0,1,2,1} Arrange the terms as shown below; Learning working make money

DSP – DFT Solved Examples Example 1 Verify Parseval’s theorem of the sequence $x(n) = frac{1^n}{4}u(n)$ Solution − $displaystylesumlimits_{-infty}^infty|x_1(n)|^2 = frac{1}{2pi}int_{-pi}^{pi}|X_1(e^{jomega})|^2domega$ L.H.S $displaystylesumlimits_{-infty}^infty|x_1(n)|^2$ $= displaystylesumlimits_{-infty}^{infty}x(n)x^*(n)$ $= displaystylesumlimits_{-infty}^infty(frac{1}{4})^{2n}u(n) = frac{1}{1-frac{1}{16}} = frac{16}{15}$ R.H.S. $X(e^{jomega}) = frac{1}{1-frac{1}{4}e-jomega} = frac{1}{1-0.25cos omega+j0.25sin omega}$ $Longleftrightarrow X^*(e^{jomega}) = frac{1}{1-0.25cos omega-j0.25sin omega}$ Calculating, $X(e^{jomega}).X^*(e^{jomega})$ $= frac{1}{(1-0.25cos omega)^2+(0.25sin omega)^2} = frac{1}{1.0625-0.5cos omega}$ $frac{1}{2pi}int_{-pi}^{pi}frac{1}{1.0625-0.5cos omega}domega$ $frac{1}{2pi}int_{-pi}^{pi}frac{1}{1.0625-0.5cos omega}domega = 16/15$ We can see that, LHS = RHS.(Hence Proved) Example 2 Compute the N-point DFT of $x(n) = 3delta (n)$ Solution − We know that, $X(K) = displaystylesumlimits_{n = 0}^{N-1}x(n)e^{frac{j2Pi kn}{N}}$ $= displaystylesumlimits_{n = 0}^{N-1}3delta(n)e^{frac{j2Pi kn}{N}}$ $ = 3delta (0)times e^0 = 1$ So,$x(k) = 3,0leq kleq N-1$… Ans. Example 3 Compute the N-point DFT of $x(n) = 7(n-n_0)$ Solution − We know that, $X(K) = displaystylesumlimits_{n = 0}^{N-1}x(n)e^{frac{j2Pi kn}{N}}$ Substituting the value of x(n), $displaystylesumlimits_{n = 0}^{N-1}7delta (n-n_0)e^{-frac{j2Pi kn}{N}}$ $= e^{-kj14Pi kn_0/N}$… Ans Learning working make money

DSP – DFT Linear Filtering DFT provides an alternative approach to time domain convolution. It can be used to perform linear filtering in frequency domain. Thus,$Y(omega) = X(omega).H(omega)longleftrightarrow y(n)$. The problem in this frequency domain approach is that $Y(omega)$, $X(omega)$ and $H(omega)$ are continuous function of ω, which is not fruitful for digital computation on computers. However, DFT provides sampled version of these waveforms to solve the purpose. The advantage is that, having knowledge of faster DFT techniques likes of FFT, a computationally higher efficient algorithm can be developed for digital computer computation in comparison with time domain approach. Consider a finite duration sequence, $[x(n) = 0,quad for,n<0quad andquad ngeq L]$ (generalized equation), excites a linear filter with impulse response $[h(n) = 0,quad forn<0quad andquad ngeq M]$. $$x(n)y(n)$$ $$output = y(n) = sum_{k = 0}^{M-1}h(k).x(n-k)$$ From the convolution analysis, it is clear that, the duration of y(n) is L+M−1. In frequency domain, $$Y(omega) = X(omega).H(omega)$$ Now, $Y(omega)$ is a continuous function of ω and it is sampled at a set of discrete frequencies with number of distinct samples which must be equal to or exceeds $L+M-1$. $$DFTquad size = Ngeq L+M-1$$ With $omega = frac{2pi}{N}k$, $Y(omega) = X(k).H(k)$, where k=0,1,….,N-1 Where, X(k) and H(k) are N-point DFTs of x(n) and h(n) respectively. $x(n)& h(n)$ are padded with zeros up to the length N. It will not distort the continuous spectra $X(omega)$ and $H(omega)$. Since $Ngeq L+M-1$, N-point DFT of output sequence y(n) is sufficient to represent y(n) in frequency domain and these facts infer that the multiplication of N-point DFTs of X(k) and H(k), followed by the computation of N-point IDFT must yield y(n). This implies, N-point circular convolution of x(n) and H(n) with zero padding, equals to linear convolution of x(n) and h(n). Thus, DFT can be used for linear filtering. Caution − N should always be greater than or equal to $L+M-1$. Otherwise, aliasing effect would corrupt the output sequence. Learning working make money

DSP – Computer Aided Design FIR filters can be useful in making computer-aided design of the filters. Let us take an example and see how it works. Given below is a figure of desired filter. While doing computer designing, we break the whole continuous graph figures into discrete values. Within certain limits, we break it into either 64, 256 or 512 (and so on) number of parts having discrete magnitudes. In the above example, we have taken limits between -π to +π. We have divided it into 256 parts. The points can be represented as H(0), H(1),….up to H(256). Here, we apply IDFT algorithm and this will give us linear phase characteristics. Sometimes, we may be interested in some particular order of filter. Let us say we want to realize the above given design through 9th order filter. So, we take filter values as h0, h1, h2….h9. Mathematically, it can be shown as below $$H(e^{jomega}) = h_0+h_1e^{-jomega}+h_2e^{-2jomega}+…..+h_9e^{-9jomega}$$ Where there are large number of dislocations, we take maximum points. For example, in the above figure, there is a sudden drop of slopping between the points B and C. So, we try to take more discrete values at this point, but there is a constant slope between point C and D. There we take less number of discrete values. For designing the above filter, we go through minimization process as follows; $H(e^{jomega1}) = h_0+h_1e^{-jomega1}+h_2e^{-2jomega1}+…..+h_9e^{-9jomega1}$ $H(e^{jomega2}) = h_0+h_1e^{-jomega2}+h_2e^{-2jomega2}+…..+h_9e^{-9jomega2}$ Similarly, $(e^{jomega1000}) = h_0+h_1eH^{-jomega1000}h_2e^{-2jomega1000}+…..+h_9+e^{-9jomega1000}$ Representing the above equation in matrix form, we have − $$begin{bmatrix}H(e^{jomega_1})\.\.\H(e^{jomega_{1000}}) end{bmatrix} = begin{bmatrix}e^{-jomega_1} & … & e^{-j9omega_1} \. & & . \. & & . \e^{-jomega_{1000}} &… & e^{j9omega_{1000}} end{bmatrix}begin{bmatrix}h_0\.\.\h_9end{bmatrix}$$ Let us take the 1000×1 matrix as B, 1000×9 matrix as A and 9×1 matrix as $hat{h}$. So, for solving the above matrix, we will write $hat{h} = [A^TA]^{-1}A^{T}B$ $= [A^{*T}A]^{-1}A^{*T}B$ where A* represents the complex conjugate of the matrix A. Learning working make money

DSP – Z-Transform Properties In this chapter, we will understand the basic properties of Z-transforms. Linearity It states that when two or more individual discrete signals are multiplied by constants, their respective Z-transforms will also be multiplied by the same constants. Mathematically, $$a_1x_1(n)+a_2x_2(n) = a_1X_1(z)+a_2X_2(z)$$ Proof − We know that, $$X(Z) = sum_{n=-infty}^infty x(n)Z^{-n}$$ $= sum_{n=-infty}^infty (a_1x_1(n)+a_2x_2(n))Z^{-n}$ $= a_1sum_{n = -infty}^infty x_1(n)Z^{-n}+a_2sum_{n = -infty}^infty x_2(n)Z^{-n}$ $= a_1X_1(z)+a_2X_2(z)$ (Hence Proved) Here, the ROC is $ROC_1bigcap ROC_2$. Time Shifting Time shifting property depicts how the change in the time domain in the discrete signal will affect the Z-domain, which can be written as; $$x(n-n_0)longleftrightarrow X(Z)Z^{-n}$$ Or $x(n-1)longleftrightarrow Z^{-1}X(Z)$ Proof − Let $y(P) = X(P-K)$ $Y(z) = sum_{p = -infty}^infty y(p)Z^{-p}$ $= sum_{p = -infty}^infty (x(p-k))Z^{-p}$ Let s = p-k $= sum_{s = -infty}^infty x(s)Z^{-(s+k)}$ $= sum_{s = -infty}^infty x(s)Z^{-s}Z^{-k}$ $= Z^{-k}[sum_{s=-infty}^infty x(m)Z^{-s}]$ $= Z^{-k}X(Z)$ (Hence Proved) Here, ROC can be written as Z = 0 (p>0) or Z = ∞(p<0) Example U(n) and U(n-1) can be plotted as follows Z-transformation of U(n) cab be written as; $sum_{n = -infty}^infty [U(n)]Z^{-n} = 1$ Z-transformation of U(n-1) can be written as; $sum_{n = -infty}^infty [U(n-1)]Z^{-n} = Z^{-1}$ So here $x(n-n_0) = Z^{-n_0}X(Z)$ (Hence Proved) Time Scaling Time Scaling property tells us, what will be the Z-domain of the signal when the time is scaled in its discrete form, which can be written as; $$a^nx(n) longleftrightarrow X(a^{-1}Z)$$ Proof − Let $y(p) = a^{p}x(p)$ $Y(P) = sum_{p=-infty}^infty y(p)Z^{-p}$ $= sum_{p=-infty}^infty a^px(p)Z^{-p}$ $= sum_{p=-infty}^infty x(p)[a^{-1}Z]^{-p}$ $= X(a^{-1}Z)$(Hence proved) ROC: = Mod(ar1) < Mod(Z) < Mod(ar2) where Mod = Modulus Example Let us determine the Z-transformation of $x(n) = a^n cos omega n$ using Time scaling property. Solution − We already know that the Z-transformation of the signal $cos (omega n)$ is given by − $$sum_{n=-infty}^infty(cos omega n)Z^{-n} = (Z^2-Z cos omega)/(Z^2-2Zcos omega +1)$$ Now, applying Time scaling property, the Z-transformation of $a^n cos omega n$ can be written as; $sum_{n=-infty}^infty(a^ncos omega n)Z^{-n} = X(a^{-1}Z)$ $= [(a^{-1}Z)^2-(a^{-1}Z cos omega n)]/((a^{-1}Z)^2-2(a^{-1}Z cos omega n)+1)$ $= Z(Z-a cos omega)/(Z^2-2az cos omega+a^2)$ Successive Differentiation Successive Differentiation property shows that Z-transform will take place when we differentiate the discrete signal in time domain, with respect to time. This is shown as below. $$frac{dx(n)}{dn} = (1-Z^{-1})X(Z)$$ Proof − Consider the LHS of the equation − $frac{dx(n)}{dn}$ $$= frac{[x(n)-x(n-1)]}{[n-(n-1)]}$$ $= x(n)-X(n-1)$ $= x(Z)-Z^{-1}x(Z)$ $= (1-Z^{-1})x(Z)$ (Hence Proved) ROC: R1< Mod (Z) <R2 Example Let us find the Z-transform of a signal given by $x(n) = n^2u(n)$ By property we can write $Zz[nU(n)] = -Zfrac{dZ[U(n)]}{dz}$ $= -Zfrac{d[frac{Z}{Z-1}]}{dZ}$ $= Z/((Z-1)^2$ $= y(let)$ Now, Z[n.y] can be found out by again applying the property, $Z(n,y) = -Zfrac{dy}{dz}$ $= -Zfrac{d[Z/(Z-1)^3]}{dz}$ $= Z(Z+1)/(Z-1)^2$ Convolution This depicts the change in Z-domain of the system when a convolution takes place in the discrete signal form, which can be written as − $x_1(n)*x_2(n) longleftrightarrow X_1(Z).X_2(Z)$ Proof − $X(Z) = sum_{n = -infty}^infty x(n)Z^{-n}$ $= sum_{n=-infty}^infty[sum_{k = -infty}^infty x_1(k)x_2(n-k)]Z^{-n}$ $= sum_{k = -infty}^infty x_1(k)[sum_n^infty x_2(n-k)Z^{-n}]$ $= sum_{k = -infty}^infty x_1(k)[sum_{n = -infty}^infty x_2(n-k)Z^{-(n-k)}Z^{-k}]$ Let n-k = l, then the above equation cab be written as − $X(Z) = sum_{k = -infty}^infty x_1(k)[Z^{-k}sum_{l=-infty}^infty x_2(l)Z^{-l}]$ $= sum_{k = -infty}^infty x_1(k)X_2(Z)Z^{-k}$ $= X_2(Z)sum_{k = -infty}^infty x_1(Z)Z^{-k}$ $= X_1(Z).X_2(Z)$ (Hence Proved) ROC:$ROCbigcap ROC2$ Example Let us find the convolution given by two signals $x_1(n) = lbrace 3,-2,2rbrace$ …(eq. 1) $x_2(n) = lbrace 2,0leq 4quad andquad 0quad elsewhererbrace$ …(eq. 2) Z-transformation of the first equation can be written as; $sum_{n = -infty}^infty x_1(n)Z^{-n}$ $= 3-2Z^{-1}+2Z^{-2}$ Z-transformation of the second signal can be written as; $sum_{n = -infty}^infty x_2(n)Z^{-n}$ $= 2+2Z^{-1}+2Z^{-2}+2Z^{-3}+2Z^{-4}$ So, the convolution of the above two signals is given by − $X(Z) = [x_1(Z)^*x_2(Z)]$ $= [3-2Z^{-1}+2Z^{-2}]times [2+2Z^{-1}+2Z^{-2}+2Z^{-3}+2Z^{-4}]$ $= 6+2Z^{-1}+6Z^{-2}+6Z^{-3}+…quad…quad…$ Taking the inverse Z-transformation we get, $x(n) = lbrace 6,2,6,6,6,0,4rbrace$ Initial Value Theorem If x(n) is a causal sequence, which has its Z-transformation as X(z), then the initial value theorem can be written as; $X(n)(atquad n = 0) = lim_{z to infty} X(z)$ Proof − We know that, $X(Z) = sum_{n = 0} ^infty x(n)Z^{-n}$ Expanding the above series, we get; $= X(0)Z^0+X(1)Z^{-1}+X(2)Z^{-2}+…quad…$ $= X(0)times 1+X(1)Z^{-1}+X(2)Z^{-2}+…quad…$ In the above case if Z → ∞ then $Z^{-n}rightarrow 0$ (Because n>0) Therefore, we can say; $lim_{z to infty}X(z) = X(0)$ (Hence Proved) Final Value Theorem Final Value Theorem states that if the Z-transform of a signal is represented as X(Z) and the poles are all inside the circle, then its final value is denoted as x(n) or X(∞) and can be written as − $X(infty) = lim_{n to infty}X(n) = lim_{z to 1}[X(Z)(1-Z^{-1})]$ Conditions − It is applicable only for causal systems. $X(Z)(1-Z^{-1})$ should have poles inside the unit circle in Z-plane. Proof − We know that $Z^+[x(n+1)-x(n)] = lim_{k to infty}sum_{n=0}^kZ^{-n}[x(n+1)-x(n)]$ $Rightarrow Z^+[x(n+1)]-Z^+[x(n)] = lim_{k to infty}sum_{n=0}^kZ^{-n}[x(n+1)-x(n)]$ $Rightarrow Z[X(Z)^+-x(0)]-X(Z)^+ = lim_{k to infty}sum_{n = 0}^kZ^{-n}[x(n+1)-x(n)]$ Here, we can apply advanced property of one-sided Z-Transformation. So, the above equation can be re-written as; $Z^+[x(n+1)] = Z[X(2)^+-x(0)Z^0] = Z[X(Z)^+-x(0)]$ Now putting z = 1 in the above equation, we can expand the above equation − $lim_{k to infty}{[x(1)-x(0)+x(6)-x(1)+x(3)-x(2)+…quad…quad…+x(x+1)-x(k)]}$ This can be formulated as; $X(infty) = lim_{n to infty}X(n) = lim_{z to 1}[X(Z)(1-Z^{-1})]$(Hence Proved) Example Let us find the Initial and Final value of x(n) whose signal is given by $X(Z) = 2+3Z^{-1}+4Z^{-2}$ Solution − Let us first, find the initial value of the signal by applying the theorem $x(0) = lim_{z to infty}X(Z)$ $= lim_{z to infty}[2+3Z^{-1}+4Z^{-2}]$ $= 2+(frac{3}{infty})+(frac{4}{infty}) = 2$ Now let us find the Final value of signal applying the theorem $x(infty) = lim_{z to infty}[(1-Z^{-1})X(Z)]$ $= lim_{z to infty}[(1-Z^{-1})(2+3Z^{-1}+4Z^{-2})]$ $= lim_{z to infty}[2+Z^{-1}+Z^{-2}-4Z^{-3}]$ $= 2+1+1-4 = 0$ Some other properties of Z-transform are listed below − Differentiation in Frequency It gives the change in Z-domain of the signal, when its discrete signal is differentiated with respect to time. $nx(n)longleftrightarrow -Zfrac{dX(z)}{dz}$ Its ROC can be written as; $r_2< Mod(Z)< r_1$ Example Let us find the value of x(n) through Differentiation in frequency, whose discrete signal in Z-domain is given by $x(n)longleftrightarrow X(Z) = log(1+aZ^{-1})$ By property, we can

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