ω = 1/R 1 C 1, which is called the 3dB cut off frequency.

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1 Lab 2 Linear Networks and Modular Construction Networks which contain only inductor, capacitors and resistors are called linear networks This means that if A and B are two sine wave inputs then the output is given as in fig21 Useful networks are band pass filters and high and low pass filters (1) High Pass Filter See fig22 for the high pass circuit Chose R 1 = 100Ω, C 1 =015µF, then ω = 1/R 1 C 1, which is called the 3dB cut off frequency Procedure: (i) Construct the circuit of fig22 (ii) Choose a reasonable number of frequencies (eg 10) that cover the range of the frequency generator and apply this signal to the input V in Simultaneously apply V in to channel 1 on the oscilloscope (iii) Our goal is to measure the frequency response of the filter Connect the output V out to channel 2 of the oscilloscope Fill in a table such as the one below and plot the frequency response V out /V in as a function of ω Table Period f ω V in V out V out /V in (2) Low Pass Filter Repeat the steps above for the low pass filter in fig23 Plot the frequency response of the low pass filter on the same graph as the high pass filter Use R 2 = 1000Ω, C 2 = 0005 µf, then ω =? (3) Band Pass Filter Repeat the measurements for the band pass filter in fig24, keeping the values of the resistors and capacitors you used in the earlier circuits Plot the frequency response of the band pass filter on the same graph as the low and high pass filter Questions: (1) Let R L, R H, R T be the response functions of the low, high and band pass

2 filters Does knowing the response function of the low and high pass filters enable you to predict the response of the band pass filter? That is, in a modular design the response of the band pass filter would be R M = R L R H R T (2) Compute Z eq for the portion of the circuit from point A looking into the input module to the left (ie the high pass filter) (3) Comput the impedance Z L of the low pass filter (ie looking into the circuit to the right from point A) (4) What is the ratio Z L /Z eq at f = 3000cps? Modular Construction When designing or analyzing circuits with some degree of complexity it is often of great convenience if one is able to consider separate sections of the circuit as modules independent of other sections of the circuit In principle, no portion of a circuit is truly independent of the other sections However, this idealized independence is approachable under certain very general conditions Consider fig25 If the output impedance of module A ( ie the equivalent impedance of module A) is very much smaller than the input impedance of a subsequent module B, then connecting the output of A to the input of B will have only a small effect on the output voltage signal of module A First let us suppose that the output of A, when module B is not connected is V A When module B s input is connected at point A 0 the signal will no longer be V A but rather V A This is easy to see once we realize that the two impedances Z eq and Z in form a voltage divider V A = V eq and V A = Z in Z in +Z eq V A Hence, it is clear if we want to minimize the effect of module B (called the loading) on the signal from A we need to make Z eq Z in As a concrete example, let s compute the true response R T of the circuit in fig24 taking account of the loading of the high pass filter by the low pass filter

3 when the connection is made at point A We redraw fig24 as in fig26 for clarity The impedance measured from point A to ground is We let the impedances of capacitors 1,2 be represented by Z 1,2 = j/(ωc 1,2 ) so V out = R 2 + V A, and, V A = V out = R 2 + V out V in = R 2 + V in, so V in, thus the true response is We note that R L = R 2 + is the complex response of the low pass filter and = R 1(R 2 + ) R 1 +R 2 +, dividing numerator and denominator of this expression by gives = R 1(R 2 + )/ R 1 Z + R = R 1/R L R 1 / +1/R L R 1 R L R 1 +Z 1 + Z 1 R 1 R L = R L (1+Z 1 /R 1 )+ Z 1 R L, or R HR L, but Z 1+ Z 1 1 Z R H R L = jωc 2 jωc 1 so 2 R HR L 1+ C 2 C 1 R H R L Hence if the ratio C 2 /C 1 is very small then the true response will approximately be, R M = R H R L R T We evaluate R H and R L R H = 1 1 j/ωτ 1, τ 1 = R 1 C 1 R L = 1 1 jωτ 2, τ 2 = R 2 C 2 Since we actually measure V out / V in we see that R M = ((1 + τ 2 /τ 1 ) 2 + (ωτ 2 1/(ωτ 1 )) 2 )) 1/2 and finally R M 1+( C 2 C ) 2 R M 2 +2 C 2 1 C (1+ τ 2 1 τ1 ) R M 2 R T = Compute the response expected for the bandpass filter using this result and

4 compare it to that obtained from the simple modular formula R H R L Plot these two response on the same graph Modularity is not automatic It is possible to install a different low pass filter at point A in fig 24 which has exactly the same filtering properties as the original low pass filter but which has a very different input impedance For example, suppose the low pass parameters are changed to C 2 = 015µF and R 2 = 33Ω This gives the same time constant τ 2 = C 2 R 2 as the original low pass filter Leave the high pass filter parameters as given and compute and plot the response functions for this new band pass filter

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