The simulation diagram of a discrete time system is shown below Find the first s

Question

The simulation diagram of a discrete time system is shown below Find the first six output (y(0) to y(6)) of the system when an input x(n), as computed in problem 3. is applied to the discrete time system. Given the transfer function shown below, convert it into a difference equation The impulse response of a discrete time system is given by To such a system we apply an input of the type Use MATLAB to convolve the two sequences and enter the answer below A system is described by the transfer function below. Find the location of poles and zeros of the system and write them here. Is this a stable system? Why or why not? The transfer function of a discrete time system is given by The system has a sampling rate of 100 Hz. Calculate the frequency response of the system at 25 Hz input frequency as a complex number in both rectangular and polar form. Hint: First calculate ohm using the relationship ohm = 2pi f/fs. where f is the applied frequency and fs is the sampling frequency (note that 0 is in radians) As pointed out in the lecture, replace Z by e(f) in the transfer function, substitute the numerical value of O. and then use Euler's formula to find the result as a complex number

Explanation / Answer

for the 4th question input to the system is not mentioned, but this can be solved easily.

the Z-1 blocks appering tin the diagram are nothing but delay blocks (i.e.,the out put of the block is delayed by one unit of time with respect to input)

assuming input x[n]= (1,1,1,1,1,1,1) n=0 to 6

output can be given by y[n] = x[n] + 2*x[n-1] + 0.5*y[n-1] + 0.5*y[n-2]

y[0] = 1

y[1] = 3.5

y[2] = 4.75

y[3]=7.125

y[4]=8.9375

y[5]=11.03125

y[6]=12.984125

a) the difference equation represented by above diagram is

y[n] = x[n] + 2*x[n-1] + 0.5*y[n-1] + 0.5*y[n-2]

5) convolution of above given to sequence = [2 -1 5 3 1 6]

6) zeros is where H(z) = 0 and poles is where H(z) = infinity

therefore; poles = i*0.8

zeros = -0.9+0.68*i and -0.9-0.68*i

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