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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)
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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 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 n0

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:

2

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 n0 as u[n] =1 in that range AND 0,otherwise.

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

(iv)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.

(vi)x[n]*(delta [n+1]+delta [n-1])

first we will see what is delta [n]:

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

so delta [n+1] = 1 at n+1=0 that is at n=-1 and delta [n-1] = 1 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

(i) 2-3*x[n]

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.

(iii)3*x[n-2]

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...}

(v)1+ 2*x[n-2]

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!

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