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Right option is (a) Multiplication in z-DOMAIN

Easy explanation: This is OBSERVED from the fact that a signal can be represented as a summation of IMPULSES and while we perform convolution in a TIME domain, we can represent the entire OUTPUT in the form of z^-n.

The correct OPTION is (a) t<0

To explain I WOULD say: Since the system is CAUSAL, the impulse response won’t exist for t<0. This is because the output of the system should not depend on future INPUTS. Thus only option t<0 is correct.

Right answer is (a) True

The explanation: The THREE delays received by the output of the transfer function suggests that the response of the transfer function is a PARABOLA itself. THUS the above STATEMENT is necessary for this CASE.

Right answer is (a) Multiplication by z^-1

The best explanation: If a signal is delayed by one unit in time, the Z-transform of the RESULTANT signal is the Z-transform of the ORIGINAL signal which gets MULTIPLIED by z^-1. HENCE, the correct option is Multiplication by z^-1.

Right answer is (a) Sampling Period

The best I can explain: This EQUATION is used to TRANSFORM the signal from Laplacian DOMAIN to z domain. Here, T REFERS to the sampling period since the entire signal needs to be SAMPLED at a period of T to be expressed in the z-domain.

The correct ANSWER is (a) A RAMP FUNCTION

The explanation: The transfer function is a step function and the step response of a step function is a ramp function. HENCE, the output in the scope is just a ramp function.

The correct answer is (a) -∞ to ∞

Easiest EXPLANATION: Unilateral Z-Transform RANGES are provided in 0 to ∞ and -∞ to 0. For BILATERAL Z-transform the signal can be DEFINED in the range given in option -∞ to ∞.

Right option is (a) The entire Z-plane

The explanation: The impulse FUNCTION has a Z-transform EQUAL to 1.Since it is independent of z, it exists for all values of z. Hence, the Z-transform CONVERGES for all values of z. Thus, the R.O.C. of impulse function is the entire z-plane.

Right answer is (a) An attenuated SIGNAL of same frequency

The explanation is: The above transfer function is that of a high PASS filter. The cut-off frequency for allowing signals is 3Hz but the given signal frequency is that of 2Hz only. If the frequency of the signal GENERATOR was more than 2Hz- the output would’ve been entire signal at the same frequency but for >>3Hz. Now, the convolution in TIME domain is multiplication in frequency domain and the output of the transfer function block is the product of the transfer function and the laplace transform of the sinusoid. It can be checked in MATLAB that the poles of the resultant function will consist of ONE pole at s=-2+(-).707i which equivalently suggest that there is a 3 db decade if the input frequency becomes less than 2hz.

The CORRECT choice is (a) INTERIOR

The explanation: For σ<0, |z| is LESS than 1. Hence, the point lies interior to the CIRCLE |z|=1.

Correct option is (a) A unit STEP signal delayed by 2 UNITS in TIME

For explanation I WOULD say: The DELAY block will delay the signal by 2 units in time.

The correct ANSWER is (b) DEL(t-100)

Easy explanation: The above graph SHOWS the impulse FUNCTION at t=+100. This means that the function has been delayed by 100 units of time. Hence, the correct option is del(t-100).

Right ANSWER is (a) True

The best I can explain: The CONTINUOUS SIGNAL can be represented as an integral of impulses. This representation buries it down to the form of convolution of two signal where ONE signal is the impulse FUNCTION while the other is the continuous signal. Hence, the above statement is true.

Right option is (a) CAUSAL

For explanation I would say: 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.

The correct choice is (a) True

For explanation: ‘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.

Correct option is (d) 2 UNITS

The best I can explain: If we observe the process of graphical convolution, we find that since, the given two signals are equal to each other, the maximum amplitude of the signal generating DUE to convolution will be midway between the total LENGTH of the resultant signal. This is due to the fact that at that INSTANT, one signal lies on top of ANOTHER. Now the total length of the convolved signal is 5 units(0,1,2,3,4) and the middle point is t=2. Hence, the correct option is t=2 units.

Correct OPTION is (a) Squaring the signal

Easiest explanation: This is proved by the fact that since DISCRETE signals can be thought of as a one VARIABLE polynomial with the COEFFICIENTS, along with the order, representing the amplitude, at an instant EQUAL to the order of the variable, of the signal- they are simply multiplied during convolution.

Correct answer is (b) u[n-3]

For explanation I WOULD say: SINCE the step time is 1, the INPUT step signal is delayed by 1 UNIT already. MOREOVER, multiplying the signal by z-2 delays the signal further by 2 units.

Right CHOICE is (a) TRUE

To EXPLAIN I would say: The signal needs to be a SAMPLED SEQUENCE so that it can be represented in terms of the complex frequency z. Hence, the above statement is true.

The correct choice is (c) A delay in the function by 3 units in time

To explain I WOULD SAY: The CONVOLUTION of an impulse function with a function results in the function itself. But if the impulse function is delayed, the output will also get delayed by an equal amount. This is because

∫^∞-∞ F(k).δ(t-m-k)dk=f(t-m).

Right ANSWER is (a) True

Explanation: The Z-transform of the impulse response of DISCRETE time system gives the transfer function of the system in z-domain. This can also be achieved from the impulse invariance transformation. HENCE, the CONVERSE should be true and the above statement is true.

Right ANSWER is (a) True

Best explanation: The associative property of convolution SUGGESTS that convolving a and B is same as convolving b and a. HENCE, the above statement is true.

Correct choice is (a) |Z|>|1|

Easiest explanation: The unit STEP function is a causal INFINITE duration signal. The R.O.C. of a^nu[n] is |z|>|a|. For a unit step function, a=1 and thus the R.O.C. is GIVEN by option only.

The CORRECT option is (a) 2-d convolution

For explanation: The conv2() COMMAND is defined in MATLAB to PERFORM 2d convolution. Hence, only 2-d convolution is correct. This TYPE of convolution is essential in image processing and VIDEO processing.

Correct OPTION is (a) Impulse INVARIANT transformation

Explanation: The Z-TRANSFORM of a signal at z=e^sT YIELDS the Laplace transform of the signal. This method of transformation is called Impulse Invariant Transformation.

The correct answer is (a) 1 and 3

Explanation: In scope 1, the INPUT of the Transfer function block is U(t-2) and the step response leads to multiplication in z-domain which shows that the ramp output GETS advanced by 1 unit and the RESULT of scope1 is r(t-1). After going through Delay of z-2, the output is r(t-3).

The CORRECT answer is (B) False

The explanation: 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.

Correct answer is (b) False

To explain I would say: The Z-transform does follow the PRINCIPLES of HOMOGENEITY and superposition. Hence, the linearity principle can be APPLIED to check if a system is linear or not in the z-domain.

Correct choice is (B) r(t)-r(t-50)-u(t-50)

To elaborate: A RAMP function extends upto infinity. But here, the graph ends a 50 units in time. So, the function has a slope change at t=50 and the new slope is 0. But the function value is 0 FO r t>50 so we need to SUBTRACT the function, resulting at t=50, from a unit step function starting from t=50. Hence, the correct solution is option c only.

Correct answer is (b) u(t)-u(t-1000)

To explain I would SAY: The unit step SIGNAL extends from 0 to infinity. When we plot it in MATLAB, we bound it to a CERTAIN point in time. In this CASE, we have bounded it to 1000 units in time. HENCE, option u(t)-u(t-1000) is correct. Option a, c and d mean the same thing but it’s not possible to represent the unit step function graphically.