Integral square error computation

[brad sue]

Hi,
I am struggling computing the integral square error between two square waves. one is ideal and the other is assymetric.
the equation for the assymetric is:
x(t)=
A for |t|<=To/4+ε/2
and
-A for (To/4)+(ε/2)<|t|<=To/2

we were asked to plot the ideal( without the ε) and actual (or assymetric) waveforms and compute the integral square error between the two waveforms.

I only have the final solution which is 4εA^2. I cannot find it
Please can someone help me?

Thank you
B

 

[mighty2000]

renda

Hi Brenda,

I can definitely help you with computing the integral square error between these two square waves. First, let's define the ideal square wave as x1(t) and the asymmetric wave as x2(t). We can write the equations for these waves as:

x1(t) = A for |t| <= To/4
x2(t) = A for |t| <= To/4 + ε/2 and x2(t) = -A for (To/4) + (ε/2) < |t| <= To/2

To compute the integral square error, we need to integrate the squared difference between x1(t) and x2(t) over the time interval from -To/2 to To/2. This can be written as:

ISE = ∫[x1(t) - x2(t)]^2 dt from -To/2 to To/2

Since x1(t) and x2(t) are both square waves, we can simplify this to:

ISE = ∫(A - x2(t))^2 dt from -To/2 to To/2

Next, we need to break this integral into two parts, one for the first half of the wave (from -To/2 to 0) and one for the second half (from 0 to To/2). This is because the waveforms are different in each half. So our integral becomes:

ISE = ∫(A - x2(t))^2 dt from -To/2 to 0 + ∫(A - x2(t))^2 dt from 0 to To/2

Now, we can substitute the equations for x2(t) into each integral:

ISE = ∫(A - A)^2 dt from -To/2 to 0 + ∫(A - (-A))^2 dt from 0 to To/2
ISE = ∫(0)^2 dt from -To/2 to 0 + ∫(2A)^2 dt from 0 to To/2

Simplifying further, we get:

ISE = 0 + 4A^2 ∫dt from 0 to To/2
ISE = 0 + 4A^2 (To/2 - 0)
ISE = 2A^2To

Now, we can see that the ISE depends on the amplitude and period of the square wave.