EC Signals and Systems

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1 UNIT I CLASSIFICATION OF SIGNALS AND SYSTEMS Continuous time signals (CT signals), discrete time signals (DT signals) Step, Ramp, Pulse, Impulse, Exponential 1. Define Unit Impulse Signal [M/J 1], [M/J 9] t δ(τ)dτ = ; t 1; t + 2. Sketch the given signal x[n] = u[n] u[n 2]. [M/J 1] 1 u[n] 1 u[n-2] > n u[n]-u[n-2] > n > n 3. Prove that the complex exponential signal x(t) = e jω t is periodic and that the fundamental period is 2π ω. [M/J 11], [N/D 9], [M/J 9], [N/D 7] The condition for periodic signal is x(t)=x(t+t) e jω t = e jω (t+t) e jω t = e jω t e jω T Department of Electronics and Communication Engineering 127

2 e jωt = e jωt (cosω T + jsinω T) e jωt = e jωt [cos(2πft) + jsin(2πft)] e jωt = e jωt [cos(2π) + jsin(2π)] f = 1/T e jωt = e jω t [cos(2π) + jsin(2π)] = 1 Hence the given complex exponential signal x(t) = e jω t is periodic. The fundamental period is ω T = 2π T = 2π ω 4. If the discrete time signal is x[n] = {,,, 3, 2, 1, 1, 7, 6}, determine y[n] = x[2n 3]. [N/D 7] 1 x[n] 1 x[n-3] > n x[2n-3] > n > n Classification of CT and DT signals periodic and non-periodic, random signals 5. Determine whether the given signal is energy or power signal and calculate the energy or power of x(t) = e 2t u(t). [N/D 12] Given signal x(t) = e 2t u(t) The energy of the signal is E = lim T T T T x 1 (t) 2 dt = lim T e 4t dt = 1 4 e 4t 1 = [ 1] 4 = 1 4 The power of the signal Department of Electronics and Communication Engineering 128

3 1 P = lim T 2T x(t) 2 dt 1 = lim T Department of Electronics and Communication Engineering 129 T T T 2T e 4t dt = lim T ( 1 8T ) e 4T = lim 8T = The energy of the signal is finite and power is zero. Therefore the given signal x(t) is an energy signal. T 1 6. Determine the even and odd components of the signal x(t) = e jt. [N/D 8] The even component of the signal x(t) is x(t) + x( t) x e (t) = 2 For the given signal x(t) = e jt, the even component is x e (t) = ejt +e jt = cos t For the given signal x(t) = e jt, the odd component is 2 x o (t) = x(t) x( t) 2 = ejt e jt 2 = jsin t CT systems and DT systems, Basic properties of systems 7. When is a system said to be memoryless? [M/J 1] A system is said to be memoryless if its output for each value of the independent variable is dependent only on the input at that same time. For example, the following system equation (1) is memoryless, as the value of y[n] at particular time n depends only on the value of x[n] at that time. y[n] = (2x[n] x[n]) 2. (1) 8. Prove that the system classified by y(t) = e x(t) is time invariant or not. [M/J 7] A system is said to be time-invariant if its input-output characteristics do not change with time. Given signal is y(t) = e x(t) To check this system is time invariant, it is necessary to determine whether the time invariance property holds for any input and any time shift t. Let x 1 (t) be an arbitrary input to this system and y 1 (t) = e x 1(t) (1) be the corresponding output. Then consider a second input obtained by shifting x1(t) in time: x 2 (t) = x 1 (t t ) (2) The output corresponding to this input is

4 y 2 (t) = e x 2(t) = e x 1(t t ) (3) The delayed output from equation (1) is y 1 (t t ) = e x 1(t t ) (4) Comparing equation (3) and (4), y 2 (t) = y 1 (t t ), (input-output characteristics do not change with time). Hence this system is time invariant. 9. Write any two properties of discrete time systems. [N/D 6] The properties of discrete time systems are i. Stability ii. Linearity 1. What are the conditions for a system to be LTI system? [N/D 13] The conditions for a system to be LTI system are, i. the system must possesses Linearity and ii. time invariant property Linearity: An LTI system, is a system that possesses the important property of superposition, that is the output of the system to a weighted sum of inputs is equal to the weighted sum of the outputs corresponding to each of the individual inputs. Time Invariant: A system is said to be time-invariant if its input-output characteristics do not change with time. Department of Electronics and Communication Engineering 13

5 UNIT II ANALYSIS OF CONTINUOUS TIME SIGNALS Fourier series analysis 1. What are the Dirichlet s conditions of Fourier series? [N/D 6], [M/J 9] The Dirichlet s conditions are Condition i: Over any period x(t) must be absolutely integrable, that is i. x(t) dt < T (1) Condition ii: In any finite interval of time, x(t) is of bounded variation, that is, there are no more than a finite number of maxima and minima during any single period of the signal. Condition iii: In any finite interval of time there are only finite numbers of discontinuities. Furthermore, each of these discontinuities must be finite 2. Define Fourier Series [N/D 9] Any periodic signal can be represented as an infinite sum of the sine and cosine functions which themselves are periodic signals of angular frequencies,, Ω,., kω. This series of sine and cosine terms are known as trignometric Fourier series and can be written as x(t) = a + [a n cos (nω t + b n sin(nω t))] n=1 where a n,b n are constants a is coefficient, also called dc component. 3. Determine the Fourier series coefficients for the signal cos(πt). [M/J 12] The Fourier series coefficients for the signal cos(πt) is determined by simply expand the cosine function as a linear combination of complex exponentials and identify by inspection of Fourier series coefficients. The cos(πt) can be expressed as cos(πt) = ejπt +e jπt (1) In general x(t) can be represented as x(t) = a k e jkω t k= (2) Compare equation number (1) and (2) possible values of k are k=1, k=-1 and k 1 or 1 a 1 = 1 2, a 1 = 1 2 and a k =, k 1 or 1 Department of Electronics and Communication Engineering 131

6 Spectrum of CT signals 4. Define Nyquist Rate [M/J 12] The sampling frequency ω s is also referred to as the Nyquist frequency. The frequency 2ω M, which under the sampling theorem must be exceeded by the sampling frequency, is commonly referred to as the Nyquist rate. ω s > 2ω M Fourier transform in signal analysis 5. Write any two properties of continuous time Fourier transform. [M/J 1] Linearity property If x 1 (t) F.T X 1 (ω) and x 2 (t) F.T X 2 (ω) then ax 1 (t) + bx 2 (t) F.T ax 1 (ω) + bx 2 (ω) Time shifting property If x(t) F.T X(ω) then x(t t ) FT e jω t X(ω) 6. Write the synthesis and analysis equations of continuous time Fourier transform. [N/D 12] Synthesis equation of continuous time Fourier transform x(t) = 1 2π X(ω)ejωt dω Analysis equation of continuous time Fourier transform X(ω) = x(t)e jωt dt 7. Derive the Fourier transform of x(t) = e at sin(ωt)u(t). [M/J 9] The Fourier transform of x(t) is X(ω) = x(t)e jωt dt X(ω) = e at sin(ωt) e jωt u(t)dt Department of Electronics and Communication Engineering 132

7 X(ω) = ( ejωt e jωt ) e jωt e at dt 2j X(ω) = 1 2j (e at e j2ωt e at )dt X(ω) = 1 2j {[e at a ] X(ω) = 1 2j {[e at a ] X(ω) = [ e (j2ω+a) (j2ω + a) ] [ e (j2ω+a) (j2ω + a) ] ω a 2 + j2aω 8. Derive the Laplace transform of the signal x(t) = e at u(t). [M/J 1] The Laplace transform of the signal x(t) X(S) = x(t)e st dt X(S) = e at u(t)e st dt X(S) = e (s+a)t dt X(S) = [ e (s+a) (s + a) ] 1 X(S) = (s + a) [e e ] 1 X(S) = [ 1] (s + a) 1 X(S) = (s + a) 9. State the initial and final value theorems of Laplace transform. [N/D 12] The initial value theorem states that x( +) = lim sx(s) s While the final value theorem states that lim t x(t) = lim s sx(s) } } Department of Electronics and Communication Engineering 133

8 1. Determine the Laplace transform of x(t) = e at sin(ωt)u(t). [M/J 9] Given w.k.t x(t) = e at sin(ωt)u(t) e at sin(ω t) u(t) L.T The Laplace transform of given x(t) is X(S) = ω (s + a) 2 + ω 2 ω (s + a) 2 + ω 2 Department of Electronics and Communication Engineering 134

9 UNIT III LINEAR TIME INVARIANT - CONTINUOUS TIME SYSTEM Convolution integral 1. State the convolution integral for continuous time LTI systems. The convolution integral y(t) of two signals x(t) and h(t) is y(t) = [M/J 1], [M/J 11] x(τ)h(t τ)dτ. The output y(t)of the system can be obtained using convolution integral if the input of the system h(t) are known. The convolution integral is given by y(t) = = x(τ)h(t τ)dτ 2. What are the four steps to obtain convolution? [M/J 9] Department of Electronics and Communication Engineering 135 i. Graph the signals x(τ) and h(τ) as a function of independent variable τ. ii. Obtain the signal h(t- τ) by folding h(τ) about τ= and then time shifting by time t. iii. Graph both signals x(τ) and h(t- τ) together on the same τ axis beginning with very large negative time shift t. iv. Multiply the 2 signals x(τ) and h(t- τ) and integrate over the overlapping interval of two signals to obtain y(t). v. Increase the time shift t and take the new interval whenever the function of either x(τ) or h(t- τ) changes, and calculate y(t) for this new interval using Step iv. vi. Repeat step v and iv for all intervals. Impulse response 3. What is the impulse response of two LTI system connected in parallel? [M/J 1] If the two systems whose impulse responses h 1 (t) and h 2 (t) are in parallel, then the overall impulse response is h(t) = h 1 (t) + h 2 (t). 4. Prove that the causality of the system with impulse response h(t) = e t u(t). [N/D 12] A system is said to be causal if the output at time t, depends on the present and past inputs but not future inputs. These systems are called non anticipative systems. h(t) = for t < for causal system. In the given system, substitute t = -1 h(-1) = e (-1) u(-1) u(-1) = for t <. So the system is causal.

10 5. What is meant by impulse response of a system? [M/J 11] x(t) = Input of a LTI continuous time system. y(t) = Response of a LTI continuous time system. h(t) = Impulse response y(t) = x(t) * h(t) yields x(τ)h(t τ)dτ. On taking Laplace transform on both sides, L{y(t)} = Y(s) => X(s).H(s) So H(s) = Y(s) X(s) => h(t) = L -1 [ Y(s) X(s) ] Fourier and Laplace transform in analysis 6. What is the Laplace transform for the function x(t) = u(t) u(t 2)? [N/D 9] X(s) = [u(t) u(t 2)] e -st dt = u(t) e -st dt - u(t 2) e -st dt = 1 - s 2 e -st dt u(t-2) = for t < 2 = 1 s - [1 s e-st ] 2 = 1 for t 2 = 1 s - [1 s e-2s ] = 1 s e 2s s = 1 s [1- e-2s ] 7. Determine the Laplace transform of the signal δ(t 5) and u(t 5). [M/J 12] i. Laplace transform of δ(t 5) L [δ(t 5)] = e -5s X(s) X(s) = 1 for δ(t) = e -5s ii. Laplace transform of u(t 5) L [u(t 5)] = e 5s s 8. What are the transfer functions of the following systems? a) An ideal integrator b) An ideal delay of T second [N/D 9] a) An ideal integrator y(t) = t x(τ) dτ Department of Electronics and Communication Engineering 136

11 If the zero initial state is zero then, y(t) = t x(τ) dτ Applying Laplace transform yields Y(s) = X(s) s H(s) = Y(s) X(s) => 1 s b) An ideal delay of T seconds, the output is y(t) = x(t-t) Applying Laplace transform yields Y(s) = e -st X(s) H(s) = Y(s) => e -st X(s) Block diagram representation 9. List out the basic elements for the block diagram representation of the continuous time system. [N/D 12] The basic elements of block diagram are i. Input x(t) = Its Laplace transform is X(s) ii. Output y(t) = Its Laplace transform is Y(s) H(s) is the transfer function of the element. Other elements in the block diagram are Summer, Gain and Multiplier. 1. Write the condition for the LTI system to be causal and stable. [N/D 6] The condition for causality in a LTI system is : The system output should depend on present and past inputs but not future inputs. A system is said to be bounded input bounded output if and only if every bounded input produced bounded output. Department of Electronics and Communication Engineering 137

12 UNIT IV ANALYSIS OF DISCRETE TIME SIGNALS Baseband sampling 1. State sampling theorem. [M/J 7], [M/J 1], [M/J 11], [N/D 7] A continuous time signal can be completely represented in its samples and recovered back if the sampling frequency is twice of the highest frequency content of the signal. i.e.,f s 2W 2. What is an antialiasing filter? [N/D 9] A filter that is used to reject high frequency signals before it is sampled to remove the aliasing of unwanted high frequency signals is called an antialiasing filter. 3. Define Aliasing [M/J 1] Aliasing is defined as when the sampling rate f s <2W,then spectrum overlap with each other. DTFT and properties 4. Define DTFT and Inverse DTFT [N/D 12] DTFT: X (e jω ) = n= x(n) e jωn IDTFT: x (n) = 1 2π X(ejω ) e jωn dω π π 5. Write the time shifting and frequency shifting properties of DTFT. [M/J 6] Time shifting: It states that if F[x(n)]=X[e jω ] Then F[x(n-k)] =e jωk X(e jω ), where k as an integer. Frequency shifting: It states that if F[x(n)]=X[e jω ] Then F[x(n) e jω n ]=X[e j(ω ω ) ] Z-transform and properties 6. Write the convolution property of the Z-transform. [N/D 12] It states that if Z {x 1 (n)} =X 1 (Z) and Z {x 2 (n)} =X 2 (Z) Then, Z {x 1 (n)* x 2 (n)} =X 1 (Z). X 2 (Z) 7. Write the final value theorem. [M/J 12] Final value theorem of Z-transform states that if x (n) is casual Z{x (n)} =X (Z) where ROC for X (Z) includes but is not necessarily confined to Z >1 and (Z-1) X (Z) has no poles on or outside the unit circle, then Department of Electronics and Communication Engineering 138

13 x ()= lim Z 1 (Z 1). X(Z) 8. What is ROC in Z-transform? [M/J 11] If x(n) is a given signal, the values of z for which the sum of converges is called ROC x(n) = n= x(n)z n The region of converges (ROC) of X (z) attains a finite value. 9. Write Parseval s relation in Z-transform. [M/J 7], [M/J 9], [M/J 12] Parseval s relation in z-transform states that, if x1(n) and x2(n) are complex-value sequences then, n= x 1 (n). x 1 (n) = 1 2j c X 1 (v). X 2 ( 1 v ). v d 1 1. Determine the Z-transform of the given data sequence: [M/J 6] 1; < n < 1 x[n] = { ; otherwise Z[x(n)] = X(z) = n= x(n). z n 1; < n < 1, here x(n) = { ; otherwise X(z) = 1 + z 1 + z 2 + z 3 + z 4 + z 5 + z 6 + z 7 + z 8 + z Determine the Z-transform for x(n) = a n 1 u (n 1). [M/J 7] x(n) = a n 1 u(n 1) z- transform of the sequence x(n) = a n u(n) X(z) = 1 1 a.z 1 ROC: lzl > lal By using time shifting property of z- transform z{x(n k)} = z k. X(z) In the same way, z{a n 1 u(n 1)} = z 1 1 a.z 1= 1 z a ROC: lzl > lal 12. Determine the Z-transform of x(n) = sin (ω n) u(n). [M/J 12] x(n) = sin(w n) u(n) X (z) = n= x(n)z n X (z) = n= sin (w n). u (n). z n X(z)= n= sin(w n). z n X (z) = [ ejwn e jw n ]. z n n= 2 Department of Electronics and Communication Engineering 139

14 X (z) = 1 2j [ejwn e jwn ]. z n n= X (z) = 1 [ 2j ejwn. z n e jw n. z n ] n= X (z) = 1 [ 2j n= (e jw. z 1 ) n n= (e jw. z 1 ) n ] If e jw. z 1 < 1 & e jw. z 1 <1, i.e., z >1 X (z) = 1 2j [ 1 1 e jw.z e jw.z 1] X (z) = 1 2j [ 1 e jw. z 1 1+e jw. z 1 (1 e jw. z 1 )(1 e jw. z 1 ) ] X (z) = 1 2j [ e jw. z 1 e jw. z 1 ] (1 e jw. z 1 e jw.z 1 +z 2 ) X (z) = [ (sin w ). z 1 (1 2(cos w ). z 1 + z 2 ) Note: n= = 1+r+r 2 + sin θ = ejθ e jθ 2j = 1, if r < 1, 1 r ] ; ROC: z > 1 Department of Electronics and Communication Engineering 14

15 UNIT V LINEAR TIME INVARIANT DISCRETE TIME SYSTEMS Difference equation 1. Determine the transfer function of the following difference equation: [N/D 8] Soln: y(n) y(n 1) + y(n 2) = x(n) Given difference equation y[n] 3 y[n 1] + 1 y[n 2] = x[n] (1) 4 8 z-transform of equation number (1) is y(z) 3 4 y(z)z y(z)z 2 = X(z) The transfer function is Y(z) [1 3 4 z z 2 ] = X(z) H(z) = Y(z) X(z) = z z 2 2. Determine the transfer function of the following difference equation: [M/J 12] Soln y(n) 5y(n 1) = x(n) + x(n 1) Given difference equation is y[n] 5y[n 1] = x[n] + x[n 1] (1) z-transform of equation number (1) is The transfer function is Y(z) 5Y(z)z 1 = X(z) + X(z)z 1 Y(z)[1 5z 1 ] = X(z)[1 + z 1 ] H(z) = Y(z) X(z) = 1 + z 1 1 5z 1 Department of Electronics and Communication Engineering 141

16 3. Write the difference equation for non-recursive system. [M/J 11] The difference equation for non-recursive system is N M a k y[n k] = b k x[n k] k= k= M N y[n] = 1 { b a k x[n k] a k y[n k] } k= k=1 Convolution sum 4. Determine the convolution of the signals x[n] = {2, 1, 3, 2} and h(n) = {1, 1, 1, 1}. [M/J 12] Soln The convolution of thr given signal x[n] and h[n] is y[n] = x[n] h[n] y[n] = {2, 3,6,,,5,2} 5. Calculate the convolution of the following signals, a) x(n) * δ(n) b) x(n) * [h1(n) + h2(n)]. [M/J 11] Soln: a) x[n] δ[n] = x[k]δ[n k] k= b) x[n] (h 1 (n) + h 2 [n]) = x[n] h 1 [n] + x[n] h 2 [n] Department of Electronics and Communication Engineering 142

17 x[n] (h 1 [n] + h 2 [n]) = x[k]h 1 [n k] + k= x[k]h 2 [n k] k= 6. State the properties of convolution sum. [M/J 1] i. The commutative property x[n] h[n] = h[n] x[n] = ii. The Distributive property h[k]x[n k] k= x[n] (h 1 (n) + h 2 [n]) = x[n] h 1 [n] + x[n] h 2 [n] iii.the Associative property x[n] (h 1 (n) h 2 [n]) = (x[n] h 1 [n]) h 2 [n] 7. Define Convolution Sum. [N/D 13] The LTI system equation is x[n] h[n] = x[k]h[n k] k= The above equation is referred to as the convolution sum or superposition sum. The operation on the right hand side of equation is known as the convolution of the sequences x[n] and h[n]. y[n] = x[n] h[n] Block diagram representation 8. Draw the general block diagram of the parallel and cascade form structure. [N/D 6] Soln: Block diagram of the cascade form structure of H(z) Given H(z) is splitted into multiplication of two or more subsystem H(z)=H 1 (z)h 2 (z) x[n] H 1 (z) H 2 (z) y[n] Fig: Block diagram of the cascade form structure Block diagram of the parallel form structure of H(z) Department of Electronics and Communication Engineering 143

18 Given H(z) is splitted into addition of two or more subsystem H(z)=H 1 (z)+h 2 (z) x[n] H 1 (z) + y[n] H 2 (z) Fig:Block diagram of the parallel form structure of H(z) LTI systems analysis using DTFT and Z-transform 9. What are the conditions for a discrete time LTI system to be causal and stable? Conditions for a discrete-time LTI system to be causal [N/D 8] A discrete-time LTI system with rational system function H(z) is causal if and only if the ROC is the exterior of a circle outside the outermost pole Conditions for a discrete-time LTI system to be stable A discrete-time LTI system is stable if and only if the ROC of its system function H(z) includes the unit circle z = Determine the system with system function and region of convergence :, is causal or stable. [N/D 13] Soln: Given H(z) = z z 1 H(z) = z z z. The poles are z=1/2 and z=2. z 2 i) A discrete-time LTI system with rational system function H(z) is causal if and only if the region of convergence is the exterior of a circle outside the outermost pole. Department of Electronics and Communication Engineering 144

19 Here the outermost pole is z=2. For given region of convergence z < 1 (which is inside 2 the outermost pole), the system function H(z) is not causal. ii) A discrete-time LTI system is stable if and only if the region of convergence of its system function H(z) includes the unit circle z = 1. Given region of convergence z < 1 does not include the unit circle z = 1. Hence the 2 given system function H(z) is not stable. Department of Electronics and Communication Engineering 145

LTI Systems (Continuous & Discrete) - Basics

LTI Systems (Continuous & Discrete) - Basics 1. A system with an input x(t) and output y(t) is described by the relation: y(t) = t. x(t). This system is (a) linear and time-invariant (b) linear and time-varying