$i)x(t-3)\\ ii) -2x (t) \\ iii) x (t-3)\\ iv) \dfrac {dx(t)}{dt}$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2015

Let amplitude of the signal is 1.

x(t) is a rectangular pulse of amplitude 1 and duration $-1 ≤ t ≤ 1.$

$(i) x(t - 3)$

x(t - 3) is a time delayed signal. Shift x(t) to the right by 3 position. Therefore, the signal will now have the duration $2 ≤ t ≤ 4.$

$(ii) -2 x(t)$

A scalar number -2 is multiplied to the signal x(t). This will make the amplitude twice of the original amplitude. Also the signal will be upside down because of the negative sign,

$(iii) x(t - 3) – 2x(t)$ This is just the addition of the two signals.

$(iv) \dfrac {d x(t)}{dt}$

This is the derivative of the signal x(t). Since it is a constant amplitude signal, the slope of the signal is zero i.e. derivative of x(t) = 0.