Fourier series of a square wave:
We have,
ak = A sinc( K
T
T
)
Signal Analysis:
The magnitude spectrum of above signal is shown below:
Here we assume (K/T) =1 at K =4,
i.e. (/T) = ¼
Duty cycle= 25%
So the multiples of the fourth harmonics are always zero.
Now,
Let us consider the compression of the above signal by the factor  (put =2 for
simplicity).
The compressed signal is x(t). But in our particular case the compressed signal is x(2t).
After compression,
Previously, o = 2/T
But now,  = 2 = 2o
T/2
The compressed signal has a different spectrum than the normal one.
Comparison of the spectrums is shown below:
So expansion in frequency domain means compression in time domain.
When the square wave signal is reduced to impulse train,
Fig. The impulse train
ak = (1/T)  x (t) e-jwotdt
T
=(1/T)  (t) dt
T
= 1/T
, where o =2 / T
The magnitude of all the frequency components in the magnitude spectrum of the
impulse train is constant i.e.1/T. It is shown below:
Fig. Magnitude spectrum
So we can conclude that the width of pulse is inversely proportional to the bandwidth.