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II. PASSIVE FILTERS Frequency-selective or filter circuits pass to the output only those input signals that are in a desired range of frequencies (called pass band). The amplitude of signals outside this range of frequencies (called stop band) is reduced (ideally reduced to zero). Typically in these circuits, the input and output currents are kept to a small value and as such, the current transfer function is not an important parameter. The main parameter is the voltage transfer function in the frequency domain, Hv (jω) = Vo /Vi . As Hv (jω) is complex number, it has both a magnitude and a phase, filters in general introduce a phase difference between input and output signals. To minimize the number of subscripts, hereafter, we will drop subscript v of Hv . Furthermore, we concentrate on the the “open-loop” transfer functions, Hvo , and denote this simply by H(jω). The impact of loading is sperately discussed. 2.1 | H(j ω) | Low-Pass Filters An ideal low-pass filter’s transfer function is shown. The frequency between the pass- and-stop bands is called the cut-off frequency (ωc ). All of the signals with frequencies below ωc are transmitted and all other signals are stopped. In practical filters, pass and stop bands are not clearly defined, |H(jω)| varies continuously from its maximum toward zero. The cut-off frequency is, therefore, defined √ as the frequency at which |H(jω)| is reduced to 1/ 2 = 0.7 of its maximum value. This corresponds to signal power being reduced by 1/2 as P ∝ V 2 . A series RL circuit as shown acts as a low-pass filter. For no load resistance (“open-loop” transfer function), Vo can be found from the voltage divider formula: R Vi R + jωL Stop Band ωc ω | H(j ω) | Κ 0.7Κ ωc ω L Low-pass RL filters Vo = Pass Band → H(jω) = + + Vi − R Vo − R Vo 1 = = Vi R + jωL 1 + j(ωL/R) We note 1 |H(jω)| = q 1 + (ωL/R)2 ECE65 Lecture Notes (F. Najmabadi), Winter 2009 23 It is clear that |H(jω)| is maximum when denominator is smallest, i.e., ω → 0 and |H(jω)| decreases as ω is increased. Therefore, this circuit allows “low-frequency” signals to pass through while “blocking” high-frequency signals (i.e., reduces the amplitude of the voltage of the high-frequency signals). The reference to define the “low” and “high”-frequencies is the cut-off frequency: “low”-frequencies mean frequencies much lower than ωc . To find the cut-off frequency, we note that the |H(jω)|M ax = 1 occurs at ω = 0 (alternatively find d |H(jω)| /dω and set it equal to zero to find ω = 0 which maximizes |H(jω)|). Therefore, |H(jω)|max = 1 1 1 |H(jω)|ω=ωc = √ |H(jω)|max = √ 2 2 1 =√ 2 1 + (ωc L/R)2 1 q −→ ωc L 1+ R 2 =2 ωc L =1 R → Therefore, ωc = R L and H(jω) = 1 1 + jω/ωc Input Impedance: Using the definition of the input impedance, we have: Zi = Vi = jωL + R Ii The value of the input impedance depends on the frequency ω. For good voltage coupling, we need to ensure that the input impedance of this filter is much larger than the output impedance of the previous stage. Since we do not know the frequency of the input signal, we need to ensure that good voltage coupling criteria is satisfied for all frequencies (or all possible values of Zi ). As such, the minimum value of Zi is an important number. Zi is minimum when the impedance of the inductor is zero (ω → 0). Zi |min = R Output Impedance: The output impdenace can be found by “killing” the source and finding the equivalent impdenace between output terminals: L R Zo Zo = jωL k R ECE65 Lecture Notes (F. Najmabadi), Winter 2009 24 where the source resistance is ignored. Again, the value of the output impedance depends on the frequency ω. For good voltage coupling, we need to ensure that the output impedance of this filter is much smaller than the input impedance of the next stage for all frequencies, the maximum value of Zo is an important number. Zo is maximum when the impedance of the inductor is infinity (ω → ∞). Zo |max = R Bode Plots and Decibel The voltage transfer function of a two-port network (and/or the ratio of output to input powers) is usually expressed in Bel: Number of Bels = log10 Po Pi or Vo Number of Bels = 2 log10 Vi because P ∝ V 2 . Bel is a large unit and decibel (dB) is usually used: Number of decibels = Vo 20 log10 Vi or Vo Vi = dB There are several reasons why decibel notation is used: Vo 20 log10 Vi 1) Historically, the analog systems were developed first for audio equipment. Human ear “hears” the sound in a logarithmic fashion. A sound which appears to be twice as loud actually has 10 times power, etc. Decibel translates the output signal to what ear hears. 2) If several two-port network are placed in a cascade (output of one is attached to the input of the next), the overall transfer function, H, is equal to the product of all transfer functions: |H(jω)| = |H1 (jω)| × |H2 (jω)| × ... 20 log10 |H(jω)| = 20 log10 |H1 (jω)| + 20 log10 |H2 (jω)| + ... |H(jω)|dB = |H1 (jω)|dB + |H2 (jω)|dB + ... making it easier to find the overall response of the system. 3) Plot of |H(jω)|dB versus frequency has special properties that make analysis simpler. For example, the plot asymptotes to straight lines at low and high frequencies as is shown below. ECE65 Lecture Notes (F. Najmabadi), Winter 2009 25 Also, using dB definition, we see that, there is a 3 dB difference between maximum gain and gain at the cut-off frequency: 20 log |H(jωc )| − 20 log |H(jω)|max " # |H(jωc )| 1 = 20 log = 20 log √ |H(jω)|max 2 ! ≈ −3 dB Bode plots are plots of |H(jω)|dB (magnitude) and 6 H(jω) (phase) versus frequency in a semi-log format (i.e., ω axis is a log axis). Bode plots of first-order low-pass RL filters are shown below (W denotes ωc ). |H(jω)|dB 6 H(jω) At high frequencies, ω/ωc ≫ 1, 1 |H(jω)| ≈ ω/ωc → |H(jω)|dB " # 1 = 20 log(ωc ) − 20 log(ω) = 20 log ω/ωc which is a straight line with a slope of -20 dB/decade in the Bode plot. It means that if ω is increased by a factor of 10 (a decade), |H(jω)|dB changes by -20 dB. At low frequencies, ω/ωc ≪ 1, |H(jω)| ≈ 1 which is also a straight line in the Bode plot. The intersection of these two “asymptotic” values is at 1 = 1/(ω/ωc) or ω = ωc . Because of this, the cut-off frequency is also called the “corner” frequency. The behavior of the phase of H(jω) can be found by examining 6 H(jω) = − tan−1 (ω/ωc). At low frequencies, ω/ωc ≪ 1, 6 H(jω) ≈ 0 and at high frequencies, ω/ωc ≫ 1, 6 H(jω) ≈ −90◦ . At cut-off frequency, 6 H(jω) ≈ −45◦ . ECE65 Lecture Notes (F. Najmabadi), Winter 2009 26 General first-order low-pass filters As we discussed before, transfer functions characterize a two-port network. As such, it is useful to group two-port networks into families based on their voltage transfer functions. To facilitate this grouping, the convention is to simplify the voltage transfer function to a form such that the “Real” part of the denominator of H(jω) is unity (i.e., the denominator should be 1 + j · · · or 1 − j · · · ). As we will see later in this section, this grouping will also help reduce the math that we do in analyzing various circuits. The low-pass RL filter discussed before is part of the family of first-order low-pass filters (first order means that ω appears in the denominator with an exponent of 1 or −1. In general, the voltage transfer function of a first-order low-pass filter is in the form: H(jω) = K 1 + jω/ωc The maximum value of |H(jω)| = |K| is called the filter gain. Note that the exponent of ω in the denominator is +1 so that |H(jω)| decreases with frequency (thus,a low-pass filter): |K| |H(jω)| = q 1 + (ω/ωc)2 6 ω |K| tan−1 H(jω) = − K ωc For RL filter, K = 1, and ωc = R/L. Note that K can be negative, and in that case, the “minus” sign adds 180◦ phase shift to the transfer function as is denoted by |K|/K factor above. Low-pass RC filters A series RC circuit as shown also acts as a low-pass filter. For no load resistance (“open-loop” transfer function), Vo can be found from the voltage divider formula: 1/(jωC) 1 Vi = Vi R + 1/(jωC) 1 + j(ωRC) 1 H(jω) = 1 + jωRC Vo = R + Vi − + C Vo − We see that the voltage transfer function of this circuit is similar to transfer function of a general first-order low-pass filter. So, this is a low-pass filter with K = 1 and ωc = 1/RC. (Note: we identified the circuit and found the cut-off frequency without doing any math!). ECE65 Lecture Notes (F. Najmabadi), Winter 2009 27 We could, of course, do the math following the procedure in analyzing the low-pass RL filter to get the same answer. (Exercise: Show this.). Following the same procedure as for RL filters, we find input and output Impedances 1 jωC 1 Zo = R k jωC Zi = R + and Zi |min = R and Zo |max = R R Terminated RL and RC low-pass filters + Now let us examin the effect of a load on the performance of our RL and RC filters. For this example, a resistive load is considered but the analysis can be easily extended to an impedance load. For example, consider the terminated RC filter shown: + Vi C − Vo RL − From the circuit, H(jω) = 1/(jωC) k RL R′ /R Vo = = Vi R + [1/(jωC) k RL ] 1 + j(ωR′ C) with R′ = R k RL This is similar to the transfer function for unterminated RC filter but with resistance R being replaced by R′ . Therefore, ωc = 1 1 = ′ RC (R k RL )C and H(jω) = R′ /R 1 + jω/ωc We see that the impact of the load is to reduce the filter gain (K = R′ /R < 1) and to shift the cut-off frequency to a higher frequency as R′ = R k RL < R. Input Impedance: Zi = R + 1 k RL jωC Output Impedance: Zo = R k Zi |min = R 1 jωC Zo |max = R We could have arrived at the same results using the the relationship between open-loop, Ho (jω), and terminated, H(jω), transfer functions of a two-port network: H(jω) = ZL Ho (jω) = ZL + Zo RL RL + R k 1 jωC ECE65 Lecture Notes (F. Najmabadi), Winter 2009 × 1 1 + jωRC 28 (Exercise: show this.) Also, note that the output impdenace of the terminated circuit is exactly the same as the open-loop version. Furthermore, it can be seen that as long as RL ≫ Zo or RL ≫ Zo |max = R (our condition for good voltage coupling), R′ ≈ R and the terminated RC filter will look exactly like an unterminated filter – The filter gain is one, the shift in cut-off frequency disappears, and input and output resistances become the same as before. Terminated RL low-pass filters The parameters of the terminated RL filters can be found similarly: Voltage Transfer Function: H(jω) = 1 Vo = , Vi 1 + jω/ωc ωc = (R k RL )/L. Input Impedance: Zi = jωL + R k RL , Zi |min = R k RL Output Impedance: Zo = (jωL) k R, Zo |max = R Here, the impact of load is to shift the cut-off frequency to a lower value. Filter gain is not affected. Again for RL ≫ Zo or RL ≫ Zo |max = R (our condition for good voltage coupling), the shift in cut-off frequency disappears and the filter will look exactly like an unterminated filter. Exercise: Derive above equations for the transfer function and input and output impdenacess. ECE65 Lecture Notes (F. Najmabadi), Winter 2009 29 2.2 First-order high pass filters In general, the voltage transfer function of a first-order high-pass filter is in the form: K 1 − jωc /ω H(jω) = It is a first-order filter because ω appears in the denominator with an exponent of −1. It is a high-pass filter because |H| = 0 for ω = 0 and |H| is constant for high-freqnecies. √ Paramter ωc is the cut-off freqnecy of the filter (Exercise: prove that |H(jωc )| is 1/ 2 = 0.7 of |H(jω)|M ax.) The maximum value of |H(jω)| = |K| is called the filter gain. |K| |H(jω)| = q 1 + (ωc /ω)2 6 ωc |K| tan−1 H(jω) = + K ω Bode Plots of first-order high-pass filters (K = 1) are shown below. The asymptotic behavior of this class of filters is: At low frequencies, ω/ωc ≪ 1, |H(jω)| ∝ ω (a +20dB/decade line) and 6 H(jω) = 90◦ At high frequencies, ω/ωc ≫ 1, |H(jω)| ∝ 1 (a line with a slope of 0) and 6 H(jω) = 0◦ |H(jω)| 6 H(jω) ECE65 Lecture Notes (F. Najmabadi), Winter 2009 30 High-pass RC filters C A series RC circuit as shown acts as a high-pass filter. The open-loop voltage transfer function of this filter is: + + Vi R − Vo R 1 H(jω) = = = Vi R + 1/(jωC) 1 − j(1/ωRC) Vo − Therefore, this is a first-order high-pass filter with K = 1 and ωC = 1/RC. Input and output impdenaces of this filter can be found similar to the procedure used for low-pass filters: Zi = R + 1 jωC and Zi |min = R Output Impedance: Zo = R k 1 jωC and Zo |max = R Input Impedance: High-pass RL filters R A series RL circuit as shown also acts as a high-pass filter. Again, we find the open-loop tranfunction to be: ωc = R L Input Impedance: H(jω) = Vi − 1 1 − jωc /ω Zi = R + jωL and Zi|min = R Output Impedance: Zo = R k jωL and Zo |max = R + + L Vo − Exercise: Compute the voltage transfer function and input and output impdenaces of terminated RC and RL filters. ECE65 Lecture Notes (F. Najmabadi), Winter 2009 31 2.3 Band-pass filters A band pass filter allows signals with a range of frequencies (pass band) to pass through and attenuates signals with frequencies outside this range. ωl : Lower cut-off frequency; ωu : ω0 ≡ | H(j ω) | Upper cut-off frequency; √ ωl ωu : B ≡ ωu − ωl : ω0 : Q≡ B Pass Band Center frequency; Band width; ωl Quality factor. ωu ω As with practical low- and high-pass filters, upper and lower cut-off frequencies of practical band pass filter are defined as the frequencies at which the magnitude of the voltage transfer √ function is reduced by 1/ 2 (or -3 dB) from its maximum value. Second-order band-pass filters: Second-order band pass filters include two storage elements (two capacitors, two inductors, or one of each). The transfer function for a second-order band-pass filter can be written as K ω0 ω − 1 + jQ ω0 ω |K| |H(jω)| = s ω ω0 2 2 1+Q − ω0 ω H(jω) = 6 H(jω) = − |K| ω ω0 tan−1 Q − K ω0 ω The maximum value of |H(jω)| = |K| is called the filter gain. The lower and upper cut-off √ frequencies can be calculated by noting that |H(jω)|max = K, setting |H(jωc )| = K/ 2 and solving for ωc . This procedure will give two roots: ωl and ωu . 1 K K |H(jωc)| = √ |H(jω)|max = √ = s 2 2 ωc ω0 2 1 + Q2 − ω0 ωc ωc ω0 2 Q − =1 ω0 ωc ωc ω0 =0 ωc2 − ω02 ± Q 2 → ωc ω0 − Q ω0 ωc ECE65 Lecture Notes (F. Najmabadi), Winter 2009 = ±1 32 The above equation is really two quadratic equations (one with + sign in front of fraction and one with a − sign). Solving these equation we will get 4 roots (two roots per equation). Two of these four roots will be negative which are not physical as ωc > 0. The other two roots are the lower and upper cut-off frequencies (ωl and ωu , respectively): ωl = ω0 s ω0 1 − 1+ 2 4Q 2Q s ωu = ω0 1 + ω0 1 + 2 4Q 2Q Bode plots of a second-order filter is shown below. Note that as Q increases, the bandwidth of the filter become smaller and the |H(jω)| becomes more picked around ω0 . |H(jω)|db 6 H(jω) Asymptotic behavior: At low frequencies, ω/ω0 ≪ 1, |H(jω)| ∝ ω (a +20dB/decade line), and 6 H(jω) → 90◦ At high frequencies, ω/ω0 ≫ 1, |H(jω)| ∝ 1/ω (a -20dB/decade line), and 6 H(jω) → −90◦ At ω = ω0 , H(jω) = K (purely real) |H(jω)| = K (maximum filter gain), and 6 H(jω) = 0◦ . There are two ways to solve second-order filter circuits. 1) One can try to write H(jω) in the general form of a second-order filters and find Q and ω0 . Then, use the formulas above to find the lower and upper cut-off frequencies. 2) Alternatively, one can directly find the √ upper and lower cut-off frequencies and use ω0 ≡ ωl ωu to find the center frequency and B ≡ ωu − ωl to find the bandwidth, and Q ≡= ω0 /B to find the quality factor. The two examples below show the two methods. Note that one can always find ω0 and k rapidaly as H(jω0 ) is purely real and |H(jω0 )| = k ECE65 Lecture Notes (F. Najmabadi), Winter 2009 33 Series RLC Band-pass filters L Using voltage divider formula, we have H(jω) = H(jω) = Vo R = Vi R + jωL + 1/(jωC) C + + Vi − R Vo − R 1 R + j ωL − ωC There are two approaches to find filter parameters, K, ω0, ωu , and ωl . Method 1: We transform the transfer function in a form similar to general form of the transfer function for second order bandpass filters: H(jω) = K ω ω0 1 + jQ − ω0 ω Note that the denominator of the general form is in the form 1 + j . . . Therefore, we divide top and bottom of transfer function of series RLC bandpass filters by R: H(jω) = 1 ωL 1 1+j − R ωRC Comparing the above with the general form of the transfer function, we find K = 1. To find Q and ω0 , we note that the imaginary part of the denominator has two terms, one positive and one negative (or one that scales as ω and the other that scales as 1/ω) similar to the general form of transfer function of 2nd-order band-pass filters (which includes Qω/ω0 and −Qω0 /ω). Equating these similar terms we get: ωL Q L Qω = → = ω0 R ω0 R Qω0 1 1 = → Qω0 = ω ωRC RC We can solve these two equations to find: 1 ω0 = √ LC ω0 Q= = R/L s L R2 C The lower and upper cut-off frequencies can now be found from the formulas on page 33. ECE65 Lecture Notes (F. Najmabadi), Winter 2009 34 Method 2: In this method, we directly calculate the filter parameters similar to the procedure followed for general form of transfer function in page 32. Some simplifications can be made by noting: 1) At ω = ω0 , H(jω) is purely real and 2) K = H(jω = jω0 ). Starting with the transfer function for the series RLC filter: H(jω) = R 1 R + j ωL − ωC We note that the transfer function is real if coefficient of j in the denominator is exactly zero (note that this happens for ω = ω0 ), i.e., ω0 L − 1 =0 ω0 C ω0 = √ −→ 1 LC Also K = H(jω = jω0 ) = R =1 R The cut-off frequencies can then be found by setting: K 1 |H(jωc)| = √ = √ 2 2 1 ωc L − 1+ R ωc RC 2 =2 which can be solved to find ωu and ωl . Input and Output Impedance of band-pass RLC filters 1 1 Zi = jωL + +R + R = j ωL − jωC ωC Zi |min = R occurs at ω = ω0 1 Zo = jωL + jωC ! kR → Zo |max = R ECE65 Lecture Notes (F. Najmabadi), Winter 2009 35 Wide-Band Band-Pass Filters Band-pass filters can be constructed by putting a high-pass and a low-pass filter back to back as shown below. The high-pass filter sets the lower cut-off frequency and the low-pass filter sets the upper cut-off frequency of such a band-pass filter. | H (j ω) | | H (j ω) | X | H (j ω) | | H (j ω) | 2 1 1 ω c2 ω ω ω c1 An example of such a band-pass filter is two RC low-pass and high-pass filters put back to back. These filters are widely used (when appropriate, see below) instead of an RLC filter as inductors are usually bulky and take too much space on a circuit board. 2 ω l = ωc2 R2 C1 + + + C2 Vi ω ω u = ωc1 R1 V1 − − Low−Pass Vo − High−Pass In order to have good voltage coupling in the above circuit, the input impedance of the high-pass filter (actually Zi |min = R1 ) should be much larger than the output impedance of the low-pass filter (actually Zo |max = R2 ), or we should have R1 ≫ R2 . In that case we can use un-terminated transfer functions: H(jω) = H1 (jω) × H2 (jω) = ωc1 = 1/(R1 C1 ) H(jω) = 1 1 × 1 + jω/ωc2 1 − jωc1 /ω ωc2 = 1/(R2 C2 ) 1 1 = (1 + jω/ωc2)(1 − jωc1/ω) (1 + ωc1/ωc2 ) + j(ω/ωc2 − ωc1 /ω) Again, we can find the filter parameters by either of two methods above. Transforming the transfer function to a form similar to the general form (left for students) gives: K= 1 1 + ωc1 /ωc2 Q= q ωc1 /ωc2 1 + ωc1 /ωc2 ECE65 Lecture Notes (F. Najmabadi), Winter 2009 ω0 = √ ωc1 ωc2 36 One should note that the Bode plots of previous page are “asymptotic” plots. The real H(jω) differs from these asymptotic plots, for example, |H(jω)| is 3 dB lower at the cutoff frequency. A comparison of “asymptotic” Bode plots for first-order high-pass filters are given in page 30. It can be seen that |H1 (jω)| achieves its maximum value (1 in this case) only when ω/ωc1 < 1/3. Similarly for the low pass filter, |H2 (jω)| achieves its maximum value (1 in this case) only when ω/ωc2 > 3. In the band-pass filter above, if ωc2 ≫ ωc1 (i.e., ωc2 ≥ 10ωc1), the center frequency of the filter will be at least a factor of three away from both cut-off frequencies and |H(jω)| = |H1 | × |H2 | achieves its maximum value of 1. If ωc2 is not ≫ ωc1 (i.e., ωc2 < 10ωc1), H1 and H2 will not reach their maximum of 1 and the filter |H(jω)|max = |H1 | × |H2 | will be less than one. This can be seen by examining the equation of K above which is always less than 1 and approaches 1 when ωc2 ≫ ωc1. More importantly, we can never make a “narrow” band filter by putting two first-order highpass and low-pass filters back to back. When ωc2 is not ≫ ωc1 , |H(jω)|max becomes smaller than 1. Since the cut-off frequencies are located 3 dB below the maximum values, the cut-off frequencies will not be ωc1 and ωc2 (those frequencies are 3 dB lower than |H(jω)|max = 1). The lower cut-off frequency moves to a value lower than ωc1 and the upper cut-off frequency moves to a value higher than ωc2 . This can be seen by examining the quality factor of this filter at the limit of ωc2 = ωc1 Q= q ωc1 /ωc2 1 + ωc1 /ωc2 = 1 = 0.5 1+1 while our asymptotic description of previous page indicated that when ωc2 = ωc1 , band-width becomes vanishingly small and Q should become very large. Because these filters work only when ωc2 ≫ ωc1, they are called “wide-band” filters. For these wide-band filters (ωc1 ≪ ωc2 ), we find from above: K=1 Q= H(jω) = q ωc1 /ωc2 ω0 = √ ωc2 ωc1 1 1 + j(ω/ωc2 − ωc1 /ω) We then substitute for Q and ω0 in the expressions for cut-off frequencies (page 33) to get: ωu = ω0 s s 1 ω0 ω0 q 2 1+ 1 + 4Q + 1 + = 4Q2 2Q 2Q ωl = ω0 1 + 1 ω0 ω0 q 2 −1 − = 1 + 4Q 4Q2 2Q 2Q ECE65 Lecture Notes (F. Najmabadi), Winter 2009 37 Ignoring 4Q2 term compared to 1 (because Q is small),we get: √ ωc2ωc1 ω0 q = ωc2 ωu = = Q ωc1 /ωc2 For ωl , if we ignore 4Q2 term compared to 1, we will find ωl = 0. We should, therefore, expand the square root by Taylor series expansion to get the first order term: ωl ≈ ω0 1 ω0 1 + 4Q2 − 1 = × 2Q2 = ω0 Q = ωc1 2Q 2 2Q What are Wide-Band and Narrow-Band Filters? Typically, a wide-band filter is defined as a filter with ωc2 ≫ ωc1 (or ωc2 ≥ 10ωc1). In this case, Q ≤ 0.35 (prove this!). A narrow-band filter is usually defined as a filter with B ≪ ω0 (or B ≤ 0.1ω0 ). In this case, Q ≥ 10. Example: Design a band-pass filter with cut-off frequencies of 160 Hz and 8 kHz. The load for this circuit is 1 MΩ. As this is wide-band, band-pass filter (ωu /ωl = fu /fl = 50 ≫ 1), we use two low- and high-pass RC filter stages similar to circuit above. The prototype of the circuit is shown below: R2 The high-pass filter sets the lower cut-off frequency, and the 1 MΩ load sets the output impedance of this stage. Thus: Zo |max = R1 ≪ 1 MΩ ωc (High-pass) = ωl = → R1 ≤ 100 kΩ 1 = 2π × 160 R1 C1 C1 + Vi + + C2 − V1 R1 − Low−Pass Vo − High−Pass → R1 C1 = 1 × 10−3 kΩ One should choose R1 as close as possible to 100 kΩ (to make the C1 small) and R1 C1 = 1×10−3 using commercial values of resistors and capacitors. A good set here are R1 = 100 kΩ and C1 = 10 nF. The low-pass filter sets the upper cut-off frequency. The load for this component is the input resistance of the high-pass filter, Zi |min = R1 = 100 kΩ. Thus: Zo |max = R2 ≪ 100kΩ ωc (Low-pass) = ωu = → R2 ≤ 10 kΩ 1 = 2π × 8 × 103 R2 C2 ECE65 Lecture Notes (F. Najmabadi), Winter 2009 → R2 C2 = 2 × 10−5 38 As before, one should choose R2 as close as possible to 10 kΩ and R2 C2 = 2 × 10−5 using commercial values of resistors and capacitors. A good set here are R2 = 10 kΩ and C2 = 2 nF. In principle, we can switch the position of low-pass and high-pass filter stages in a wideband, band-pass filter. However, the low-pass filter is usually placed before the high-pass filter because the value of capacitors in such an arrangement will be smaller. (Try redesigning the above circuit with low-pass and high-pass filter stages switched to see that one capacitor become much smaller and one much larger.) Exercise: Design an RLC filter with the specifications in the previous example. (Hint: Do not set R = 100 kΩ as this would make the value of the inductor very large.) ECE65 Lecture Notes (F. Najmabadi), Winter 2009 39 2.4 Exercise Problems In circuit design, use 5% commercial resistor and capacitor values (1, 1.1, 1.2, 1.3, 1.5, 1.6, 1.8, 2, 2.2, 2.4, 2.7, 3., 3.3, 3.6, 3.9, 4.3, 4.7, 5.1, 5.6, 6.2, 6.8, 7.5, 8.2, 9.1 × 10n where n is an integer). Youcan also use 10 mH inductors Problem 1. Design a RLC bandpass filter with a lower cut-off frequency of 1 kHz and a bandwidth of 3 kHz. What is the center frequency and Q of this filter? Problem 2. We have an amplifier that amplifies a 1 kHz signal from a detector. The load for this amplifier can be modeled as a 50 kΩ resistor. The amplifier output has a large amount of 60 Hz noise. We need to reduce the amplitude of noise by a factor of 10. Design a first-order passive filter which can be placed between the amplifier and the load and does the job. Would this filter affect the 1 kHz signal that we are interested in? If so, by how much? Problem 3. The tuner for an FM radio requires a band-pass filter with a central frequency of 100 MHz (frequency of a FM station) and a bandwidth of 2 MHz. a) Design such a filter (Use a 1 µH inductor). b) What are its cut-off frequencies? Problem 4. A telephone line carries both voice band (0-4 kHz) and data band (25 kHz to 1 MHz). Design a filter that lets the voice band through and rejects the data band. The filter must meet the following specifications: a) For the voice band, the change in transfer function should be at most 1 dB; and b) The transfer function should be as small as possible at 25 kHz, the low end of the data band. ECE65 Lecture Notes (F. Najmabadi), Winter 2009 40 2.5 Solution to Exercise Problems Problem 1. Design a RLC bandpass filter with a lower cut-off frequency of 1 kHz and a bandwidth of 3 kHz. What is the center frequency and Q of this filter? L The circuit prototype is: For a 2nd order band-pass filter: B(Hz) = fu − fl C + + Vi fu = 1 + 3 = 4 kHz − R Vo − B(rad/s) = 2πB(Hz) = 1.88 × 104 ωu = 2πfu = 2.51 × 104 √ ω0 = ωu ωl = 1.26 × 104 B(rad/s) = ω0 Q → Q= ωl = 2πfl = 6.28 × 103 1.26 × 104 = 0.67 1.88 × 104 For the series RLC circuit: 1 1 → LC = 2 ω0 LC 1 1 C= = = 0.63 µF Lω02 10 × 10−3 × (1.26 × 104 )2 ω0 = √ Q= ω0 R/L → R ω0 = = B(rad/s) L Q R = LB = 10 × 10−3 × 1.88 × 104 = 188 Ω Therefore, using commercial values, the design values are L = 10 mH, R = 180 Ω, and C = 0.68 µF. ECE65 Lecture Notes (F. Najmabadi), Winter 2009 41 Problem 2. We have an amplifier that amplifies a 1 kHz signal from a detector. The load for this amplifier can be modeled as a 50 kΩ resistor. The amplifier output has a large amount of 60 Hz noise. We need to reduce the amplitude of noise by a factor of 10. Design a first-order passive filter which can be placed between the amplifier and the load and does the job. Would this filter affect the 1 kHz signal that we are interested in? If so, by how much? We want to have 1 kHz signals to go through but reduce 60 Hz signals, so we need a high-pass filter. The prototype of the circuit is shown below. For this circuit: C H(jω) = + + 1 Vo = Vi 1 − jωc /ω Inverting Amp. 1 RC Zi |min = R ωc = Vi R Vo RL − − Zo |max = R As the output impedance of the inverting amplifier circuit is “zero”, we do not need to worry about the input impedance of our filter. The output impedance of the filter is restricted by Zo |max = R ≪ 50 kΩ → R ≤ 5 kΩ This filter should reduce the amplitude of 60 Hz (ω60 = 2π × 60 = 120π rad/s) signal by a factor of 10, i.e., |H(jω = jω60 )| = Vo Vi 1 + (ωc /ω60 )2 = 100 60 1 = 0.1 =q Hz 1 + (ωc /ω60 )2 → 1 = ωc ≈ 10ω60 = 3751 rad/s RC → RC = 2.67 × 10−4 Reasonable choices are R = 3.9 kΩ (to keep it below 5 kΩ) and C = 68 nF (fc ≈ 600 Hz). The impact on 1 kHz signal (ω1000 = 2000π rad/s) can be found from: 1 1 =q = 0.86 |H(jω = jω1000 )| = q 1 + (ωc /ω1000 )2 1 + (3751/6283)2 So the amplitude of 1 kHz signal is reduced by 14% (or by -1.3 dB). ECE65 Lecture Notes (F. Najmabadi), Winter 2009 42 Problem 3. The tuner for an FM radio requires a band-pass filter with a central frequency of 100 MHz (frequency of a FM station) and a bandwidth of 2 MHz. a) Design such a filter (Use a 1 µH inductor). b) What are its cut-off frequencies? L Because this is not a wide-band filter, the simplest filter will be an RLC filter as is shown. For this filter: s + + Vi − 1 = 2π100 × 106 ω0 = √ LC ω0 Q= = B C R Vo − L 2π100 × 106 = = 50 R2 C 2π × 2 × 106 Using a L = 1 µH inductor: 1 = 4π 2 1016 LC 1 = 4π 2 1016 × 10− 6 C → → C = 2.5 × 10−12 F Choose: C = 2.2 pF L = 2, 500 R2 C → R2 = L 10−6 = = 182 → 2, 500C 2, 500 × 2.2 × 10−12 R = 13.5 Ω Choose: R = 13 Ω (L = 1 µH and C = 2.2 pF). To find the cut-off freqnecies, we not: B = fu − fl = 2 MHz f0 = q fu fL = 100 MHz Solution of the above two equations in two unknowns will give fl ≈ 99 MHz and fu ≈ 101 MHz. ECE65 Lecture Notes (F. Najmabadi), Winter 2009 43 Problem 4. A telephone line carries both voice band (0-4 kHz) and data band (25 kHz to 1 MHz). Design a filter that lets the voice band through and rejects the data band. The filter must meet the following specifications: a) For the voice band, the change in transfer function should be at most 1 dB; and b) The transfer function should be as small as possible at 25 kHz, the low end of the data band. We need a low-pass filter as it should allow low-frequency signals (voice band) to go through while eliminating high-frequency signals (data band). The prototype of an RC low-pass filter is shown and its transfer function is: R 1 1 H(jω) = = 1 + jω/ωc 1 + jf /fc + Vi + C Vo − The cut-off frequency of the filter is not given and it should be found from the specifications. First, we need the change in transfer function to be at most 1 dB for the frequency range of 0-4 kHz. The transfer function of filters that satisfy this constraint is the curve labeled “1” in the figure and any transfer function located to the right of this curve (such as transfer function labeled “2”). − | H(j ω) | 0 2 −1dB 1 f (kHz) 4 25 Second, the transfer function should be as small as possible at 25 kHz. This requires that we choose the cut-off frequency as small as possible. Therefore, the transfer function of our filter should be curve labeled “1” as it has the smallest possible value at 25 kHz: 20 log (|H(jf = 4 kHz)|) = −1 dB → |H(jf = 4 kHz)| = 0.891 Using the expression for H(jω), we have: 1 |H(jf = 4 kHz)| = q = 0.891 1 + (f /fc )2 f /fc = 0.509 fc = → 1 = 7.85 × 103 2πRC fc = f = 7.85 kHz 0.509 → RC = 2.0 × 10−5 Choosing C = 1 nF, we have R = 2.0 × 104 . The commercial values then are C = 1 nF and R = 20 kΩ. ECE65 Lecture Notes (F. Najmabadi), Winter 2009 44