Question

Discrete System question

9.3. The signals in Figure P9.2 are zero except as shown. (a) For the signal x,[n] of Figure P9.2(a), plot the following (i) 2 3n] (iii) 3x1n-2] (v) 12x-2n (ii 2xn (iv) 3- aIn (vi) 2 -n]-4 (b) Repeat part (a) for the signal xb[n] of Figure P9.2(b). (c) Repeat part (a) for the signal xn of Figure P9.2(c). (d) Repeat part (a) for the signal xn of Figure P9.2(d) 484 Discrete-Time Signals and Systems Chap. S 9.4. The signals in Figure P9.2 are zero except as shown. (a) For the signal x[n] of Figure P9.2(a), plot the following: (ii) x[n (iii) xdn]u[n + 2] (v) n-2] (b) Repeat part (a) for the signal xb[n] of Figure P9.2(b) (e) Repeat part (a) for the signal xIn] of Figure P9.2(c) (d) Repeat part (a) for the signal xnof Figure P9.2(d) (vi) 'ra[n](δ[n + 1] + 히n _ 1])

Answer 1

Hello

I am answering this question (9.4a part (i),(iv),(vi) AND 9.3a part (i),(iii),(v) only) by assuming x[n] (which can be ) because it is not given. I am here to explain the procedure for any discrete signal.(other parts can be done from explained parts)

I am answering this question (9.4a part (i),(iv),(vi) AND 9.3a part (i),(iii),(v) only) by assuming x[n] (which can be x_a [n], x_b [n], x_c[n], x_d [n]) because it is not given. I am here to explain the procedure for any discrete signal. (other parts can be done from explained parts) 9.4a (i)_x[-n]*u[n] u[n] is defined as 1 for n greaterthanorequalto 0 0, otherwise. So x[-n] is reflection of x[n] about y axis. If x[n] = {Ellipsis 0, 0, 1, -3, 2, 1, -0.5, 0, 0 Ellipsis} //discrete notation and highlighted point is x[0] so x[n] graph looks like: If x [n] is If x[n] = {Ellipsis 0, 0, 1, -3, 2, 1, -0.5, 0, 0 Ellipsis}, x[-n] = {Ellipsis 0, 0, -0.5, 1, 2, -3, 1, 0, 0 Ellipsis} so x[-n]*u[n] has only points for which n greaterthanorequalto as u[n] = 1 in that range AND 0, otherwise. so x[-n]*u[n] = {Ellipsis 0, 0, 0, 0, 2, -3, 1, 0, 0 Ellipsis} (i)_x[-n]*u[-2-n] x[-n] = { Ellipsis 0, 0, 0, 0, 2, -3, 1, 0, 0 Ellipsis } from above part.

9.4a

x[-n]*u[n]

u[n] is defined as

1 for

0, otherwise.

So x[-n] is reflection of x[n] about y axis.

If x[n] = {...0,0,1,-3,2,1,-0.5,0,0...} //discrete notation and highlighted point is x[0]

so x[n] graph looks like:

If x[n] = {...0,0,1,-3,2,1,-0.5,0,0...} , x[-n] = {...0,0,-0.5,1,2,-3,1,0,0...}

so x[-n]*u[n] has only points for which as u[n] =1 in that range AND 0,otherwise.

so x[-n]*u[n] = {...0,0,0,0,2,-3,1,0,0...}

x[-n]*u[-2-n]

x[-n] = {...0,0,-0.5,1,2,-3,1,0,0...} from above part.

Now we have to find u[-n-2]:

u[n] = {...,0,0,1,1,1,1.....} all ones after n=0(included)

lets take f[n] as u[-n] that is

f[n] = u[-n] = {...1,1,1,1,1,0,0...}

g[n] = f[n+2] = {...,1,1,1,0,0,0,0...}//advanced in time axis

f[n+2] = u[-n-2]; So g[n] is u[-n-2].

x[-n]*u[-n-2] = {...0,-0.5,0,0,0,0...} that is at n=-2 its -0.5 and everywhere 0.

first we will see what is :

It is 1 when n=0 and 0 everywhere in discrete time.

so at n+1=0 that is at n=-1 and at n-1=0 that is at n=1.

as x[n] = {...0,0,1,-3,2,1,-0.5,0,0...}

Given expression is { ...0,0,-3,0,1,0,0...}

9.3a

multiply every value of x[n] by -3 and add 2 to every value after

as x[n] = {...0,0,1,-3,2,1,-0.5,0,0...} , -3*x[n] = {...0,0,-3,9,-6,-3,1.5,0,0...} and add 2 to resulted one. So,

given expression is {...2,2,-1,11,-4,-1,3.5,2,2...} all 2s after and before.

x[n-2] is time delayed by 2 units version of x[n].

So x[n-2] is obtained by shifting x[n] by two units right which implies

x[n-2] = {...0,0,1,-3,2,1,-0.5,0,0...} See last 1 is at n=0 now

and 3*x[n-2] = {...0,0,3,-9,6,3,-1.5,0,0...}

This expression is similar to 3*x[n-2] in above part but multiply x[n-2] by 2 and add 1 to every value.

x[n-2] = {...0,0,1,-3,2,1,-0.5,0,0...} from above part.

1+ 2*x[n-2] = {...0,0,2,-6,4,2,-1,0,0...}+1 (for all)

= {...1,1,3,-5,5,3,0,1,1...} all ones before and after.

Hope this answer helps.

Thanks!