




Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Lecture 14 of a signals and systems course, focusing on filtering, frequency response, and types of filters such as lowpass, highpass, bandpass, and band-reject filters. The lecture covers the concept of frequency response, the relationship between input and output magnitudes, and the classification of filters based on their input/output relationship and passive or active components.
Typology: Study notes
1 / 8
This page cannot be seen from the preview
Don't miss anything!
Frequency Response The frequency response of a circuit is its steady-state response to a sinusoidal input as the frequency of the sinusoidal input varies.
V i = Vi∠θi and V o = Vo∠θo
Recall in Laplace analysis, the transfer function is obtained by taking the ratio between the output and input. The system's frequency response can be found by substituting jω for s.
TF (s = jω) = TF∠θTF
The relationship between input and output is
Vo∠θo = TF∠θTF Vi∠θi
A circuit's frequency response is a phasor relationship involving magnitude and phase..
OUR INTEREST IN FILTERING IS USUALLY CONFINED TO THE MAGNITUDE RESPONSE.
Find the frequency response of this system.
The nodal equations read:
Use Maple to find Vo(s) = TF(s) Vi(s)
restart; eqns:={v1=vi, (v2-v1)s/8+v2/(2s)-2*(v1 -v2)+v2/4=0}: soln:=solve(eqns,{v1,v2}): assign(soln); vo:=v2: TF:=vo/vi;
s (s + 16) TF := ------------- s^2 + 4 + 18 s
This is the system transfer function. To obtain the system frequency response, substitute jω for s.
x 1 2
1 i 2 1 2 2 1 2
Define control variable: V = V - V nodal equations voltage source equation: V = V
KCL at node 2: V^ - V^ + V^ - 2 (V - V ) + V = 0 (^8) 2s 4 s
Active Low Pass This lowpass filter has a break frequency (in rad/s) of 1/RfC and a dc gain of Rf/Rin.
There is also inversion since this is the inverting configuration.
Passive High Pass High pass filters pass high frequencies from input to output and attenuate low frequencies.
The capacitance impedance increases at low frequencies. By voltage division, Vo/Vin will decrease at low frequencies.
Active Band Pass Find the transfer function of this circuit.