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FILTER CIRCUITS The operation of many electronic instruments requires the amplification of a selected range of frequencies while suppressing others. The behaviour of an amplifier for a given frequency can be controlled by adding reactive elements - capacitors and inductors. Such circuits are usually called filters because they “filter” out frequencies except those lying in a selected range. The response of any electronic circuit depends to some degree on the frequency of operation. In an amplifier, for example, the gain will be constant over some range but fall off outside this range of frequencies. The frequency behaviour is called its frequency response characteristic and is usually displayed graphically. The ratio of output voltage to input voltage as it varies over a given frequency range is know as the circuit’s transfer function. add graphs Low-pass High-pass Band-pass Notch FILTERS 260ab.doc 260aa.ppt 10/05/17 4:25 PM 1 FILTERS 260ab.doc 260aa.ppt 10/05/17 4:25 PM 2 FILTERS 260ab.doc 260aa.ppt 10/05/17 4:25 PM 3 FILTERS 260ab.doc 260aa.ppt 10/05/17 4:25 PM 4 Simple RC SERIES CIRCUIT complex circuit Complex representation Potential differences Currents Impedances Z=R+jX XC = -1 / C XL = L simple circuit 1 1 Z1 vs Z1 vs 2 3 vS VSo e jt 5 Z5 4 Z3 Z2 Applied emf vs Z4 load Z3 1 Z2 and Z3 in parallel Z4 Z1 and Z4 in series Z5 Z1 Z4 1 1 Z 2 Z3 Z 2 Z3 Z 2 Z3 Currents and potential differences v i5 S i1 i4 Z5 v1 i1 Z1 v4 i4 Z4 v2 v3 v i2 2 Z2 i3 v3 Z3 Checks i1 i4 i2 i3 Power p1 v1 i1 FILTERS vS v1 v4 p2 v2 i2 p3 v3 i3 260ab.doc 260aa.ppt 10/05/17 4:25 PM 5 LOW PASS FILTER Control data 1 Z1 = R1 = 1000 R1 C2 = 1 F vs Z2 = - j / ( C2) Z3 = R3 = 1000 3 2 R3 C2 VSo = 10 V fmin = 10 Hz fmax = 104 Hz This circuit is called a low-pass filter because it passes low frequency signals with less attenuation than high frequency signals. A passive RC circuit such as this one does not have a sharp cut-off above a certain frequency, rather, the attenuation changes gradually. This frequency is known as the cut-off frequency (break-point frequency or corner frequency) fo. For very low frequencies, the capacitive reactance is so large compared to the resistance values that it is essentially an open circuit. Thus the circuit behaves as if the capacitor was removed. The circuit then acts as a simple voltage divider and the voltage gain Av is Av v3 R3 vs R1 R3 or R3 Av 20 log10 dB R1 R3 Since the capacitor has very little effect upon the circuit at low frequencies, the load voltage v3 and supply voltage vS are in phase ( ~ 0). For very high frequencies, the capacitive reactance is much smaller than R3, thus shorting it out. Therefore, R3 has no effect on the circuit and the circuit again acts as a simple voltage divider and the voltage gain Av is v Z2 j / C 1 jR1C Av 3 vs Z1 Z 2 R1 j / C 1 ( R1C ) 2 Since f is large ( = 2f) we can assume that R1C >> 1, then Av FILTERS 1 R1C and 1 Av (dB) 20log10 20log10 f constant R1C 260ab.doc 260aa.ppt 10/05/17 4:25 PM 6 Therefore, at high frequencies, the voltage gain falls off in inverse proportion to the frequency. This is shown as a straight line segment with a slope of –20 when the voltage gain Av (dB) is plotted against log10(f). In many applications, it is necessary to use multiple filter stages to give a greater fall off in gain with frequency. Since the capacitor dominates the circuit at high frequencies, the phase difference between the load voltage v3 and the supply voltage vS is ~ –90° (-/2 rad). The cut-off frequency fo is defined when the gain is Av Z3 1 1 max gain 2 2 Z1 Z3 1 Av (dB) 20log10 max gain 2 1 Av (dB) 20log10 20log10 max gain 2 Av (dB) 3 20log10 max gain fo 1 2 R1 R3 R1 R3 C2 if R3 very large then fo 1 2 1 R1 C2 This is referred to as the 3 dB point, since it is the frequency at which the voltage gain has fallen 3 dB. At this point, the power is ½ of the maximum power. For frequencies less than the cut-off frequency, the voltage gain is essentially constant and independent of frequency. Above the cut-off frequency, the voltage gain falls with a slope of –20 when the voltage gain Av (dB) is plotted against log10(f). Note, that at this point the phase angle is –45° ( rads), midway in its range from 0° to –90°. Maximum power transferred to load Z3 ZTH Z1 Z 2 Z1 Z 2 But Z2 is frequency dependent, therefore, the power delivered to the load depends on the frequency of the source. FILTERS 260ab.doc 260aa.ppt 10/05/17 4:25 PM 7 MATLAB FILE %m260ab.m %9 feb jan 01 %RC low pass filter %data vso = 10; fmin = 1; fmax= 1e4; num = 501; %control data z1 = 1000; c2 = 1e-6; z3 = 1000; %test z1t = c2t = z3t = data 1000; 2e-6; 1000; %calcualtions df = (fmax-fmin)/(num-1); f = fmin : df : fmax; w = f .* (2*pi); %control calculations xc = 1./(w.*c2); z2 = -j*xc; z4 = (z2 .* z3); z4 = z4 ./(z2 +z3); z5 = z1+z4; i5 = vso ./ z5; i1 = i5; i4 = i5; v3 = i4 .* z4; v3db = real(20*log10(v3/vso)); phase3 = angle(v3)/pi; power3 = abs((v3 .^2) /z3); %test calculations xct = 1./(w.*c2t); z2t = -j*xct; z4t = (z2t .* z3t); z4t = z4t ./(z2t +z3t); z5t = z1t+z4t; i5t = vso ./ z5t; i1t = i5t; i4t = i5t; v3t = i4t .* z4t; v3tdb = real(20*log10(v3t/vso)); phase3t = angle(v3t)/pi; power3t = abs((v3t .^2) /z3t); FILTERS 260ab.doc 260aa.ppt 10/05/17 4:25 PM 8 %graphing figure(1); plot(log10(f),v3db); hold on plot(log10(f),v3tdb,'m'); grid on; title('Low Pass Filter'); xlabel('log(f)'); ylabel('Voltage gain Av (dB)') figure(2); plot(log10(f),phase3); hold on plot(log10(f),phase3t,'m'); grid on; title('Low Pass Filter'); xlabel('log(f)'); ylabel('phase (in pi rad)'); figure(3); plot(log10(f),power3); hold on plot(log10(f),power3t,'m'); grid on; title('Low Pass Filter'); xlabel('log(f)'); ylabel('Pload (W)') Low Pass Filter -5 -10 Voltage gain Av (dB) -15 -20 -25 -30 -35 -40 -45 FILTERS 0 0.5 1 1.5 2 log(f) 2.5 260ab.doc 260aa.ppt 10/05/17 4:25 PM 3 3.5 4 9 Low Pass Filter Low Pass Filter 0 0 -0.05 -0.05 -0.1 -0.1 -0.15 phase (in pi rad) phase (in pi rad) -0.15 -0.2 -0.2 -0.25 -0.25 -0.3 -0.3-0.35 -0.35 -0.4 -0.45 -0.4 -0.5 -0.45 -0.5 0 0 0.5 0.5 1 1 1.5 1.5 2 log(f) 2.5 2 log(f) 3 2.5 3.5 3 4 3.5 4 Low Pass Filter 0.025 0.02 Pload (W) 0.015 0.01 0.005 0 0 0.5 1 1.5 2 log(f) 2.5 Control data results: R1 = R3 AV(max) = - 6 dB fo = 314 Hz AV = - 9 dB 3 3.5 4 Pmax = 0.025 W P = 0.0125 W (1/2 Pmax) high frequencies: slope AV/log(f) graph = (-26 + 17)/(3.5 – 3.0) = -18 (~-20) FILTERS 260ab.doc 260aa.ppt 10/05/17 4:25 PM 10