i "—--
! variation
Amplitude variation
Figure 4.11. The use of a double-tuned circuit provides increased dynamic range and improves the linearity by balancing the non-linearity of one LC circuit with the other. Note that the "push-pull" arrangement shown here will generate predominantly odd harmonics.
to give vfm(t) = A[sin oct + (mf sin ost) cos oct] (4.4.4)
vfm (t) = Afsin oc t +1 mf [sin(oc — os )t + sin(oc + os )t]}. (4.4.5)
It should be noted that the above approximation eliminates all the sidebands except the two closest to the carrier. The modulation is then said to be a narrow band frequency modulation (NBFM). The advantage it offers is a simple geometric representation of the FM signal as a phasor.
The term [(mf sin ost) cos oct] is a double-sideband-suppressed carrier (DSB-SC) signal. The addition of the carrier term (sin os t) would normally produce an AM signal but, in this case, the carrier is 90° out of phase with the DSB-SC signal. The difference between AM and FM is shown in the phasor diagrams in Figure 4.12.
Figure 4.12. (a) A phasor diagram for an NBFM showing the two phasors of amplitude 2Amf rotating in opposite directions at angular velocity os in quadrature to the phasor A at t = 0. (b) A phasor diagram for an AM signal showing two counter-rotating phasors of amplitude 1 kA and angular velocity os. Note that the amplitude of the resultant phasor varies from A(1 — k) to A(1 + k).
Figure 4.12. (a) A phasor diagram for an NBFM showing the two phasors of amplitude 2Amf rotating in opposite directions at angular velocity os in quadrature to the phasor A at t = 0. (b) A phasor diagram for an AM signal showing two counter-rotating phasors of amplitude 1 kA and angular velocity os. Note that the amplitude of the resultant phasor varies from A(1 — k) to A(1 + k).
Assume a coordinate system in which the phasor A sin oct rotates anticlockwise at an angular velocity oc. Suppose that the rotating phasor is used as a new reference system so that it is represented by a horizontal phasor of value A. In terms of this reference system, the term (2 Amf) sin(oc + os)t is represented by a phasor of value 2 Amf rotating in an anticlockwise direction with an angular velocity of os. Similarly, the term (1 Amf) sin(oc — os)t is represented by a phasor of equal value rotating in a clockwise direction with an angular velocity of os. At time t = 0, the component of each rotating phasor in the horizontal direction is zero. Therefore the two phasors must be parallel to each other but at right angles to the horizontal phasor A. This situation is depicted in Figure 4.12(a). The resultant is a phasor which apparently has a maximum value
and a maximum angular displacement of c = tan—!(mf). (4.4.7)
The apparent variation of the amplitude of the FM signal is due to the fact that this analysis has not taken into account the many sidebands present in FM.
For comparative purposes the AM system is treated similarly. An AM signal is represented by Equation (2.2.7):
vam(t) = A sin oct + kAcos(oc — os)t — ^cos(fflc + os)t. (4.4.9)
Again the carrier A sin oct is represented by a horizontal phasor of value A and the two sidebands by two counter-rotating phasors of value kA/2 rotating at os. At time t = 0, the components of both of the sideband in the horizontal direction are zero since they are cosine terms and A is a sine term. The phasors representing the two sidebands must be at right angles to the A phasor as shown in Figure 4.12(b). Note that one is positive and the other negative. At some other time t > 0, the phasor representing the combination of the sidebands subtracts from or adds to the phasor A. The amplitude of the resultant therefore varies from A(1 — k) to A(1 + k) but at all times it is in the horizontal position.
The conclusion is that in a narrow-band (Aoc ^ oc) frequency modulated system, the phase difference between the carrier and the DSB-SC signal is 90°. This is the basis of the Armstrong system. A block diagram of the FM transmitter is shown in Figure 4.13.
A crystal-controlled oscillator generates the carrier frequency A cos oct. This signal is fed to two blocks. The first is a 90° ^hase shift czrcwzt which, as the name suggests,
shifts the phase by 90° and therefore its output is A sin oct. The second block is a balanced modulator. The other input to the balanced modulator is, of course, the modulating signal. But according to Equation (4.2.8), mf = Doc/os. The modulating (audio) signal contains a band of frequencies from approximately 20 Hz to 15 kHz and therefore as os increases, Do must increase or the modulation index mf must decrease. However, in FM the frequency of the carrier changes proportionately to the amplitude of the modulating signal, not to the frequency of the modulating signal. In order to satisfy this condition, the amplitude of the modulating signal has to be reduced by one-half whenever the frequency doubles. This amounts to putting the signal through a low-pass filter with a slope of 6 dB/octave. An integrator with its corner frequency below the lowest modulating frequency is required. The amplitude of the modulating signal frequency band is thereby "predistorted'' by the integrator to conform to the above condition. The need for the integrator is evident from Equation (4.2.7).
If the modulating signal is B sin ost the output of the integrator will be (B/os) cos ost. This signal is fed to the balanced modulator which produces as an output the product of its two inputs, namely, (k0AB/os) sin oct cos ost, where k0 is a constant. This signal and the output of the 90° phase shifter go to the adder which gives an output vo (t) = A sin oc t +
k0 AB
The signal goes through a series of multipliers to bring the frequency up to the operating value. A power amplifier raises the signal to the proper level for radiation by the antenna.
In a practical FM transmitter, the carrier is generated at a lower frequency and the modulation process carried out. The frequency is then multiplied up to the operating value. If it is assumed that the sub-carrier frequency is again 200 kHz - a frequency at which a suitably robust crystal can be found - the required multiplication factor to place it in the middle of the commercial FM frequency band is then about 500 (29 = 512). The maximum frequency deviation for a sub-carrier at 200 kHz was calculated in Section 4.3.4 to be 150 Hz.
The ratio of the sub-carrier frequency to the maximum frequency deviation may not be the most convenient in a practical system. In order to produce a high-fidelity signal, that is, maintain a high level of linearity, it is an advantage to make the carrier frequency independent of the frequency deviation. This can be done by splitting the multiplication operation in two and inserting a mixer between them so that the carrier frequency can be changed without affecting the frequency deviation. To help explain the process, an example is given in Figure 4.14.
Suppose the sub-carrier frequency is 200 kHz and the maximum deviation is 20 Hz. To get the sub-carrier frequency up to the normal operating frequency of about 100 MHz, a cascade of frequency multipliers equal to 29 (512) will be required. Multiplication of the carrier also multiplies the frequency deviation so that the corresponding frequency deviation will be (512 x 10 Hz) = 10,240 Hz. This is much lower than the limit allowed, which is 75 kHz. On the other hand, in order to convert the 20 Hz frequency deviation into the required 75 kHz, the multiplication factor is 3750 (approximately 3 x 210 = 3072). However, multiplying the subcarrier by this factor will give a carrier frequency of 614.4 MHz, which is outside the commercial FM radio band. To correct this, the multiplication process is split into two stages and a mixer is inserted between them. The first stage of multiplication may be 96 (3 x 25). The carrier frequency is then 19.2 MHz and the corresponding frequency deviation is 1.92 kHz. The signal is now mixed with a ''local oscillator" whose output frequency is 16.2 MHz to produce a carrier frequency of 3.0 MHz but the frequency deviation remains unchanged at 1.92 kHz. The second stage of multiplication (32 = 25) produces a carrier frequency of 96.0 MHz and a frequency deviation of 61.4 kHz. This is not quite at the allowed limit but close enough for practical purposes. The ''local oscillator" signal may be derived from the 200 kHz crystal-controlled oscillator by using a suitable frequency multiplier (81 = 34).
In the block diagram of the Armstrong system given in Figure 4.14, four new components were added, namely the 90° phase shifter, the balanced modulator, the integrator, and the adder. The design of these circuit components now follow.
4.4.2.1 The 90° Phase Shift Circuit. In the circuit shown in Figure 4.15(a), the resistances RC and RE are equal. Since the collector and emitter currents are, for all practical purposes, equal, it follows that the voltages vC and vE must be equal in magnitude. But the phase angle between them is 180°. The voltages vC and vE are
represented on the phasor diagram shown in Figure 4.15(b). The voltage which appears across R and Z is (vC + vE). If the impedance Z is purely reactive (capacitive or inductive), it follows the phasors vR and vZ will always be at right angles to each other. It follows then that the point A must lie on a circle whose radius is equal to the magnitude of vC as shown. The output voltage is then vo. As frequency goes from zero to infinity, point A will trace out a semicircular locus. The output voltage vo changes its phase angle from zero to 180° but its magnitude remains constant.
It can be seen that, with the appropriate choice of R and Z at a given frequency, vo can be made to lead or lag vin by 90°. One major disadvantage of this circuit is that its load must be a very high impedance, preferably open-circuit, otherwise the magnitude of the output voltage vo will not remain constant for all frequencies. This circuit is a simple example of a class of networks known as all-pass filters.
Example 4.4.1 90° phase shift circuit. Design an all-pass circuit with a phase shift of 90° at 200 kHz. The dc supply voltage is 12 V and it may be assumed that the circuit drives a load of very high impedance. The b of the transistor is 100.
Solution. A suitable circuit for the phase shifter is shown in Figure 4.16 with the impedance Z replaced by a single capacitance C.
Since the gain is unity vC = vE and the stage will have the maximum dynamic range when the transistor is biased with VE at Vcc /4 and Vc at 3 Vcc/4, that is, at 3 and 9 V, respectively. The maximum signal voltage that can be applied to the RC series circuit is then 12 V peak-to-peak. The equivalent circuit is then as shown in Figure 4.17.
Let the peak current in the RC circuit be approximately equal to 1 mA. Then R = 4.2 kO and the reactance of the capacitor must also be 4.2 kO so that at 200 kHz, C = 190 pF.
The dc collector current in the transistor must be chosen so that the 1 mA taken by the RC circuit will have negligible effect on the transistor operation. A collector current of 10 mA will be adequate. Under quiescent conditions, the emitter voltage is 3 V Hence RE = 300 O and RC = 300 O. Since the b of the transistor is 100, the base current is 100 mA. Allowing a current 10 times the base current in the resistive chain (i.e. 1 mA) makes (R1 + R2) = 12 kO. But the voltage at the base is 3.7 V and therefore R2 = 3.7 kO and R1 = 8.3 kO. The coupling capacitor has to be chosen so that it is essentially a short-circuit at 200 kHz; a 0.1 mF capacitor is adequate.
4.4.2.2 The 90° Phase Shift Circuit: Operational Amplifier Version.
When the frequency of operation is within the bandwidth of an operational amplifier, the circuit shown in Figure 4.18 can be used to realize a 90° phase shift.
The analysis of the circuit is simplified by the application of superposition. The first step is to separate the inverting input from the non-inverting input. With the non-inverting input connected to ground, and vin connected to the inverting input, it can be shown that the output is
With the inverting input grounded and vin applied to the non-inverting input, it can be shown that the output is
The output when vin is connected to both inputs is v0 = v01 + v02
Was this article helpful?
Read how to maintain and repair any desktop and laptop computer. This Ebook has articles with photos and videos that show detailed step by step pc repair and maintenance procedures. There are many links to online videos that explain how you can build, maintain, speed up, clean, and repair your computer yourself. Put the money that you were going to pay the PC Tech in your own pocket.