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Textbook by: Carlos E. Davila. E-mail the author

# The Sinusoidal Steady State Response

Module by: Carlos E. Davila. E-mail the author

It is useful to see what the effect of the filter is on a sinusoidal signal, say x(t)=cos(Ω0t)x(t)=cos(Ω0t). If y(t)y(t) is the output of the filter, then we can write

y ( t ) = - cos ( Ω 0 ( t - τ ) ) h ( τ ) d τ y ( t ) = - cos ( Ω 0 ( t - τ ) ) h ( τ ) d τ
(1)

Using the Euler formula for cos(Ω0t)cos(Ω0t), right hand side of Equation 1 can be written as:

1 2 - e j ( Ω 0 ( t - τ ) ) + e - j ( Ω 0 ( t - τ ) ) h ( τ ) d τ 1 2 - e j ( Ω 0 ( t - τ ) ) + e - j ( Ω 0 ( t - τ ) ) h ( τ ) d τ
(2)

This integral can be split into two separate integrals, and written as:

e j Ω 0 t 2 - e - j Ω 0 τ h ( τ ) d τ + e - j Ω 0 t 2 - e j Ω 0 τ h ( τ ) d τ e j Ω 0 t 2 - e - j Ω 0 τ h ( τ ) d τ + e - j Ω 0 t 2 - e j Ω 0 τ h ( τ ) d τ
(3)

The first of the two integrals can be recognizes as the Fourier Transform of the impulse response evaluated at Ω=Ω0Ω=Ω0. The second integral is just the complex conjugate of the first integral. Therefore Equation 3 can be written as:

e j Ω 0 t 2 H ( j Ω 0 ) + e - j Ω 0 t 2 H * ( j Ω 0 ) e j Ω 0 t 2 H ( j Ω 0 ) + e - j Ω 0 t 2 H * ( j Ω 0 )
(4)

Since the second term in Equation 4 is the complex conjugate of the first term, we can express Equation 4 as:

R e e j Ω 0 t H ( j Ω 0 ) R e e j Ω 0 t H ( j Ω 0 )
(5)

or expressing H(jΩ0)H(jΩ0) in terms of polar coordinates:

R e e j Ω 0 t | H ( j Ω 0 ) | e j H ( j Ω 0 ) = R e | H ( j Ω 0 ) | e j ( Ω 0 t + H ( j Ω 0 ) ) R e e j Ω 0 t | H ( j Ω 0 ) | e j H ( j Ω 0 ) = R e | H ( j Ω 0 ) | e j ( Ω 0 t + H ( j Ω 0 ) )
(6)

Therefore, we find that the filter output is given by

y ( t ) = | H ( j Ω 0 ) | cos ( Ω 0 t + H ( j Ω 0 ) ) y ( t ) = | H ( j Ω 0 ) | cos ( Ω 0 t + H ( j Ω 0 ) )
(7)

This is called the sinusoidal steady state response. It tells us that when the input to a linear, time-invariant filter is a cosine, the filter output is a cosine whose amplitude has been scaled by |H(jΩ0)||H(jΩ0)| and that has been phase shifted by H(jΩ0)H(jΩ0). The same result applies to an input that is an arbitrarily phase shifted cosine (e.g. a sine wave).

Example 3.1 Find the output of a filter whose impulse response is h(t)=e-5tu(t)h(t)=e-5tu(t) and whose input is given by x(t)=cos(2t)x(t)=cos(2t). It can be readily seen that the frequency response of the filter is

H ( j Ω ) = 1 5 + j Ω H ( j Ω ) = 1 5 + j Ω
(8)

and therefore H(j2)=0.1857H(j2)=0.1857 and H(j2)=-0.3805H(j2)=-0.3805. Therefore, using Equation 7:

y ( t ) = 0 . 1857 cos ( 2 t - 0 . 3805 ) y ( t ) = 0 . 1857 cos ( 2 t - 0 . 3805 )
(9)

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