{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lec7 - AE 383 System Dynamics Lecture Notes 7(July 8th 2006...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 AE 383 System Dynamics Lecture Notes 7 (July 8 th 2006) Frequency Response (Sinusoidal Response) of 1 st Order Systems If we make our system input sinusoidal with amplitude q A and frequency ω rad /s, the system equation becomes t Kq q dt dq A ω τ sin 0 0 = + Remember that for frequency response we are interested only in the sinusoidal steady state, which is achieved after transients die out; that is we want the forced, not the natural, part of the response. We can use the method of undetermined coefficients to find the particular solution. Let q op = Asin ω t + Bcos ω t Then t Kq t B t A t B t A A ω ω ω ω ω ω ω τ sin cos sin ) sin cos ( = + + - A Kq A B = + - ω τ 0 = + B A τ ω 1 1 2 2 2 2 + - = + = τ ω ωτ τ ω A A Kq B Kq A so t Kq t Kq q A A op ω τ ω ωτ ω τ ω cos 1 sin 1 2 2 2 2 + - + = This may be simplified by using the trigonometric identity ) tan sin( cos sin 1 2 2 A B B A B A - + + = + α α α Using this, we finally have [ ] ) ( tan sin 1 1 2 2 ωτ ω τ ω - + + = - t Kq q A op Now let’s use the sinusoidal transfer function to find the same result, and see that this method is more simple 1 ) ( ) ( + = s K s q s q i o τ substituting j ω for s
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 ) ( tan 1 1 ) ( 1 2 2 0 ωτ τ ω ωτ ω - + = + = - K j K j q q i so the amplitude ratio = 1 2 2 + τ ω K phase angle ) ( tan 1 ωτ φ - = -
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}