# Solve this small problem

Discussion in 'Homework Help' started by salmanshaheen_88, Jun 24, 2010.

1. ### salmanshaheen_88 Thread Starter Active Member

Mar 5, 2009
88
1
plz solve the encircled part

2. ### tskaggs New Member

Jun 17, 2010
26
3
Part e -

$[x(t)+x(-t)]*u(t)=x(t)*u(t)+x(-t)*u(t)$

The x(t)*u(t) signal is the part of x(t) that is in the RHP.
The x(-t)*u(t) signal is the part of x(t) that is in the LHP, reversed into the RHP.
So, the sum of those to signals should be x(t)=0 until t=0. Then a straight line at amplitude 3 from t=0 to 1. And from t=1 to 2 you have a straight line from 1 to -1. The rest of time is x(t)=0.
Part f -

$x(t)*[\delta(t+\frac{3}{2})-\delta(t-\frac{3}{2})]=x(t)*\delta(t+\frac{3}{2})-x(t)*\delta(t-\frac{3}{2})$

The delta function is basically a sampler. So what you have is a positive sample of x(t) at t=-3/2 and a negative sample of x(t) at t=3/2.

3. ### salmanshaheen_88 Thread Starter Active Member

Mar 5, 2009
88
1

so how I will solve these equations??? I solved the first four parts and don't have any concept of remaining two so please solve completely

4. ### tskaggs New Member

Jun 17, 2010
26
3
They're not really equations that you solve, but descriptions of signals. For example, u(t) describes the unit step signal which is defined as the signal whose amplitude is zero at time ranging from negative infinity to 0 and 1 at time ranging from 0 to positive infinity.

I have attached a pdf that shows drawings of what I am trying to say.

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5. ### tskaggs New Member

Jun 17, 2010
26
3
I just realized that on part f I drew the samples backward. At t=-3/2, the signal should have amplitude of -1/2, and at t=3/2 the signal should have amplitude of 1/2. Sorry about that.

6. ### salmanshaheen_88 Thread Starter Active Member

Mar 5, 2009
88
1
thanks member

7. ### salmanshaheen_88 Thread Starter Active Member

Mar 5, 2009
88
1
why the value of unit step u(t) is 1 at t>0 ???

Jun 17, 2010
26
3
9. ### Papabravo Expert

Feb 24, 2006
10,711
1,992
It is so by definition. In order to analyze complex waveforms it helps to consider them as being made up of a sum of simple waveforms. The reason they are so useful in linear systems is that in a linear system you can use the principle of superposition to decompose things into simpler forms, solve them, and recombine the results.

10. ### salmanshaheen_88 Thread Starter Active Member

Mar 5, 2009
88
1
so if the qustion is like [x(2t) + x(-2t)] u(t) so still the value of unit step will be 1???

11. ### tskaggs New Member

Jun 17, 2010
26
3
Like Papabravo was saying... it is a definition. X(2t) is a signal, X(-2t) is a signal and u(t) is a signal. The resultant signal from adding, time distorting, and multiplying does not change the signal itself.