MATLAB Question Paper on “Operations on Signals – 2”.
1. What is the output of the following code?
plot([-5:1:5],tripuls([-4:1:4])
a) Error
b) Generates a triangular pulse signal
c) Generates a sawtooth signal
d) Cannot be determined
Answer: a
Clarification: The length of the first vector given as an input to the plot command is not equal to the length of the second vector given as an input. Hence, the above code will result in an error. If the first vector was ranging from [-4:1:4], the output would’ve been Generates a triangular pulse signal only.
2. What is the output of the following code?
b={1 2 3}; c={ 1 2 3} b+c
a) Error
b) { 2 4 6 }
c) { 1 2 3 }
d) Cannot be determined
Answer: a
Clarification: Mathematical operators like ‘+’, ‘-‘ do not work on cellular arrays. Hence, the above code will lead to an error in MATLAB.
3. What is the output of the following code?
t=[0:1:2]; plot(t,p); hold; plot(-t,p);
a) A mirror image of ramp function from 0 to 2 units in time
b) A ramp function from 0 to 2 units in time and it’s a mirror image in the same window
c) A ramp function from 0 to 2 units in time
d) Error
Answer: b
Clarification: Since we have used the hold command, we can now plot multiple graphs in the same window. The 1st plot command generates a ramp function from 0 to 2 units in time while the 2nd ramp function generates its mirror image in the same window. There is no error in the above code.
4. Discrete time convolution is not possible in MATLAB.
a) True
b) False
Answer: b
Clarification: We need to define the signal in terms of vectors and then we can give them as inputs to the conv() command. This will result in the discrete time convolution.
5. plot([0:1:8],[rectpulse(1,9)]’) and plot([0:1:8],ones(1,9)) will generate the same graph.
a) True
b) False
Answer: a
Clarification: ‘rectpulse(1,9)’ generates a column vector of 9 1’s. Since we have transposed it, the resultant is a row vector of 9 1s. ‘ones(1,9)’ also generates 9 1’s as a row vector. Thus, the above codes will result in the same graph.
6. If x(t-5)=0 for t<5 = A for t>=5, the signal is ___________
a) Causal
b) Symmetric
c) Linear, Time Invariant and Causal
d) Cannot be determined
Answer: a
Clarification: The signal given to us is delayed by 5 units in time. If we advance the signal by 5 units, the signal amplitude is 0 for t<0 but A for t>=0. Hence, the signal will be causal.
7. What is the output of the following code?
>>p=[zeros(1,5),ones(1,6)]; >> t=0:1:10; >> plot(t,p);
a) r(t-5)-r(t-6)-u(t-10)
b) r(t-5)-u(t-6)-u(t-10)
c) r(t-5)-u(t-6)-r(t-10)
d) r(t-5)-r(t-6)-r(t-10)
Answer: a
Clarification: The signal changes from 0 to 1 from t=5 to t=6 seconds. This indicates a ramp function starting from t=5 seconds. But again it remains constant till t=10 so it signifies that the same signal is subtracted by a ramp function at t=6 seconds. Finally the signal ends at t=10 seconds so a unit step signal gets subtracted from the previous signal at t=10 seconds.
8. What is the output of the following code?
[l,m]=deconv([4 6 8],[2 3 4])
a) l=2,m=0
b) m=2,l=0
c) l=m=2
d) Error