Manchester Encoding Matched Filter

[Meteora]

Homework Statement

Consider one bit time of Manchester code of duration T where a positive pulse of amplitude A/2 and length T/2 is followed by a negative pulse of amplitude –A/2 and length T/2.
a) Determine the impulse response of a filter matched to this signal and sketch it as a function of time.
b) Plot the matched filter output as a function of time.
c) What is the peak value of the output?

Homework Equations

Matched filter is filter with equation g(T-t) where g(t) is the input to the filter as far as I know for the Non return to zero case.

The Attempt at a Solution

I know I have to use Schwarz's Inequality and somehow get to the mathced filter representation but I can't figure out how since I only know how to do it with NonReturntoZero encoding.

 

[KataKoniK]

Hello,

Thank you for your post. To determine the impulse response of a filter matched to this signal, we can use the definition of a matched filter as a filter with a response that is the time-reversed and conjugate of the transmitted signal. In this case, our transmitted signal is a Manchester code with a positive pulse of amplitude A/2 and length T/2 followed by a negative pulse of amplitude -A/2 and length T/2.

So, the impulse response of our matched filter can be represented as g(T-t) where g(t) is the input to the filter. In this case, our input is the Manchester code, so g(t) can be written as:

g(t) = (A/2)u(t) - (A/2)u(t-T/2) - (A/2)u(t-T) + (A/2)u(t-3T/2)

where u(t) is the unit step function.

Using Schwarz's Inequality, we can simplify this expression to:

g(t) = (A/2)u(t) - (A/2)u(t-T/2) - (A/2)u(t-T) + (A/2)u(t-3T/2)
= (A/2)[u(t) - u(t-T/2) - u(t-T) + u(t-3T/2)]

Now, to sketch the impulse response as a function of time, we can plot the above expression for different values of t. For t < 0, the unit step functions will be equal to 0, so g(t) = 0. For 0 < t < T/2, u(t) = 1 and the other unit step functions are equal to 0, so g(t) = A/2. For T/2 < t < T, u(t-T/2) = 1 and the other unit step functions are equal to 0, so g(t) = 0. For T < t < 3T/2, u(t-T) = 1 and the other unit step functions are equal to 0, so g(t) = -A/2. For 3T/2 < t, u(t-3T/2) = 1 and the other unit step functions are equal to 0, so g(t) = 0.