The Dirac delta functional d(t) can only be simplified under integration. Let's see what happens to the scaling of the time axis of the Dirac delta functional under integration. First, let a > 0:
/ oo | | d(a t) dt | / -ooWe make the change of variables u = a t. Hence, du = a dt; i.e., dt = (1/a) du. Limits of integration remain the same.
/ oo / oo | 1 | 1 | d(a t) dt = - | d(u) du = - | a | a / -oo / -ooSecond, let a < 0. As before, we make the change of variables u = a t. Hence, du = a dt; i.e., dt = (1/a) du. Limits of integration reverse because a < 0.
/ oo / -oo | 1 | 1 | d(a t) dt = - | d(u) du = - - | a | a / -oo / ooTherefore, the area of the Dirac delta d(a t) is 1 / |a| instead of 1, provided that a ≠ 0.
Consider a Fourier transform of the form
G(w) = A(w) + j B(w)
where A(w) and B(w) have real-valued amplitude. Our goal is to convert this Cartesian form into polar form to be able to plot magnitude and phase responses in the frequency domain:
j Theta(w) G(w) = | G(w) | e
2 2 | G(w) | = sqrt( A (w) + B (w) ) -1 B(w) Theta(w) = tan ( ---- ) A(w)
The expression for Theta(w) requires A(w) ≠ 0. When A(w) = 0, we could use the more general two-argument arctangent function (atan2 in Matlab). Or, we could rewrite the expression for G(w) directly, as done next.
In this problem, A(w) = 0. When B(w) ≥ 0, G(w) = j B(w) can be written as
|G(w)| = B(w) j pi/2 Theta(w) = ePlease note that j = exp(j pi/2).
When B(w) < 0, G(w) = j B(w) can be written as
|G(w)| = -B(w) -j pi/2 Theta(w) = e
Please note that -j = exp(-j pi/2).
One could also solve this problem in the frequency domain. Given input x(t) and output y(t) of a linear time-invariant system with an impulse response h(t),
y(t) = h(t) * x(t)In the Fourier domain,
Y(w) = H(w) X(w)Hence, we can compute X(w) and Y(w), solve for H(w), and take the inverse Fourier transform of H(w) to get h(t).
The Fourier-domain approach is algorithmic. However, there are a few caveats: