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Signals and Systems
LTI Systems
Discrete Convolution

Questions mapped to Discrete Convolution under LTI Systems.

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Q#1 LTI Systems GATE EE 2025 (Set 1) MCQ +1 mark -0.33 marks

Consider a discrete-time linear time-invariant (LTI) system , where

 

Let

 

where  is the discrete-time unit impulse function. For an input signal , the output  is

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Q#2 LTI Systems GATE EE 2021 (Set 1) MCQ +1 mark -0.33 marks

Two discrete-time linear time-invariant systems with impulse responses  and  are connected in cascade, where  is the Kronecker delta. The impulse response of the cascaded system is

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Q#3 LTI Systems GATE EE 2017 (Set 1) NAT +2 marks -0 marks

Consider a causal and stable LTI system with rational transfer function H(z), whose corresponding impulse response begins at n = 0. Furthermore,

. The poles of H (z) are  for k = 1, 2, 3, 4.

The zeros of H(z) are all at z = 0. Let . The value of g[8] equals ___________________. (Given the answer up to three decimal places).  

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Q#4 LTI Systems GATE EE 2017 (Set 2) MCQ +2 marks -0.66 marks

A cascade system having the impulse responses and is shown in the figure below, where symbol denotes the time origin.

Z:\PY\EE\Redreaw figure\Signal & System\updated\Signal & system\145-26.jpg

The input sequence x(n) for which the cascade system produces an output sequence  is

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Q#5 LTI Systems GATE EE 2012 (Set 1) MCQ +2 marks -0.66 marks

Let y[n] denote the convolution of h[n] and g[n], where  and g[n] is a casual sequence. If  and , then g[1] equals

0

1

3/2

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Q#6 LTI Systems GATE EE 2010 (Set 1) MCQ +2 marks -0.66 marks

Given the finite length input x[n] and the corresponding finite length output y[n] of an LTI system as shown below, the impulse response h[n] of the system is  

55.jpg

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Q#7 LTI Systems GATE EE 2006 (Set 1) MCQ +2 marks -0.66 marks

x[n] = 0; n <-1&n>0, x[-1]=-1, x[0]=2 is the input and y[n]=0; n<-1&n>2, y[-1]=–1=y[1], y[0]=3, y[2]=-2 is the output of a discrete time LTI system.  The system impulse response h[n] will be

h[n]=0; n <0,n> 2,h[0] = 1,h [1] = h[2] = –1

h[n]=0; n <–1,n>1, h[–1] = 1, h[0] = h[1] = 2

h[n] = 0; n <0,n> 3, h [0] = –1, h[1] = 2,h[2] = 1

h[n] = 0; n <–2,n>1, h[-2] = h[1] = h[–1] = –h[0] = 3

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