2- a Superheterodyne Receiver Has a Minimum Detectable Signal of 0.09 Μv Into 50 Ohms

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2- A superheterodyne receiver has a minimum detectable signal of 0.09 µV into 50 ohms. The RF amplifier has gain of 15 dB and the mixer has an input third-order intecept of +12 dBm. Determine the spurious-free dynamic range at the input of the receiver.

2- A şağıdaki RF alici devresinin a) Giri şine Si = –90 dBm ve Ni = –144 dBm uygulandı ğında çıkı ştaki S o/N o ne olacaktır? b) Toplam IIP3 nedir? c) Giri şine ayni güçte iki sinyal uygulandı ğını varsayalım. Bu devrenin çıkısındaki intermodulasyon sinyalinin P imd = –90 dBm olması için giri ş sinyallerinin gücü ne olmalıdır?

2- Consider the receiver front end below.

G1=−1dB G 2 = 13 dB G 3 = 6 dB GG 45 =− 3 dB = 40dB

F1=1.26 F 2 = 2 dB F 3 = 6 dB FF 45 == 2 6dB IIP3=− 7 dBm IIP 3 = 10 dBm IIP 3 = 10 dBm

4 G1 = 0.79 G2 =19.95 G3 = 3.98 G4 = 0.5 G5 =10 F1 = 1.26 F2 = 1.58 F3 = 3.98 F4 = 2 F5 = 3.98 IIP3 = 0.2 mW IIP3 = 10 mW IIP3 = 10 mW OIP3 = 3.98 mW OIP3 = 39.81 mW OIP3 = 10 5 mW

Find S o/N o when we input S i = –90 dBm and N i = –144 dBm.

Fcas = 2.29 = 3.6 dB, S o/N o(dB) = S i/N i(dB) – F(dB) = –90 – (–144) – 3.6 = 50.4 dB.

Find the cascaded IIP3 of the whole receiver.

1 1 1 111 = + + ++ OIP3cas GGGGOIP2345 3 1 GGGOIP 345 3 2 GGOIP 45 333 3 GOIP 5 4 OIP 5 1 1 11 = + + OIP3cas GGGOIP345 3 2 GGOIP 45 33 3 OIP 5 11 11 1 =4− 3 + 4 − 353 +=→= − 0.0275OIP 3cas W OIP 3cas (4)(0.5)(10)4× 10 (0.5)(10)40 × 10 10 × 10 0.0275

OIP 3cas = 36364 mW = 45.6dB→IIP 3cas (dB) = OIP 3 cas (dB) − G cas (dB) =−− 45.6 55 = 9.4 dB c) Assume that there are two input sources to the receiver with equal power levels. How much input power will be required to produce a 3 rd order intermodulation component of –90 dBm at the output?

IIP3cas = P in + ½ (P out – P imd ). P out = P in + Gca s. Gcas = 55 dB. Pin = 2/3 (IIP3cas – ½ Gcas + ½ P imd ) = 2/3 (–9.4 – ½ (55) + ½ (–90) = –54.6 dB.

7. Consider the following RF front end.

(a) Calculate the overall NF and IIP3 of the RF front end shown. NF d = 25 dB NF c = L 3 + NF d = 29 dB 2 2 Fb = F 2 + (F c –1 ) / [(Av2) /4 =Ap2] = 2.0 + (794.33 –1 ) / [(7.94) /4] = 52.29 NF b = 17.18 dB NF a = L1 + NF b = 19.18 dB NF tot = NF a = 19.18 dB

IIP3d = 45 dBm = 56.23 Vpk = 39.76 Vrms (given) IIP3c = IIP3d + L 3 = 45 dBm + 4 dBm = 49 dBm

Since the image reject filter is a passive device we can assume that its IIP3 is very large (infinity) so that it won’t effect the overall IIP3. However the overall IIP3 is increased due to its finite loss.

2 2 2 1/IIP3b = (1 / IIP3 LNA ) + (Av 2) / (IIP3 c) = (1 / 10.1 Vpk) 2 + (7.9433) 2 / (89.119 Vpk) 2 Ï IIP3b = 0.1 Vpk = –10 dBm

IIP3a = IIP3b + L1 = –10 dBm + 2 dBm = –8 dBm IIP3tot = –8 dBm

(b) What is the sensitivity and Spurious Free Dynamic Range (SFDR) of this RF front end? Assume a 100 kHz BW and a minimum acceptable signal to noise ratio of 10 dB.

BW = 100 kHz, SNRout-min = 10 dB Assuming conjugate matching at input

Pin,min = 10log(kT) + NF + 10logB + SNRmin = –174 dBm/Hz + 19.18 dB + 10log(10 5) + 10 dB = –94.8 dBm Therefore sensitivity is –94.8159 dBm

F = –174 dBm + NF + 10logB = –174 dBm + 19.18 + 10log(10 5) = –104.82 dBm

SFDR = (2IIP3 + F)/3 – (F + SNRmin) = 2(–8 dBm +104.82)/3 – 10 dB = 54.54 dB

3- For the system below, determine the maximum input power, Pin, so that the output power signal, Pout, is at the (output) 1-dB compression point of the second amplifier.

4- Consider the following system. Calculate the overall OIP3 and IIP3.

G=10 dB G=0 dB OIP3=0 dBm OIP3=20 dBm

1 1 =1 + = 1.01dBm ⇒ OIP3= 1 = 0dBm OIP3 100

OIP3 IIP3= = 0.1 =− 10 dBm 10

6- a) Bir kuvvetlendiricinin OIP3 kesim noktasi 25 dBm ve kazanci 12 dB olsun. Cikista gozlenen intermodulasyon isaretinin seviyesi –40 dBm ise giristeki isaretin seviyesi ne olmalidir?

Pin = 2/3 (OIP3 – 3/2 G + ½ Pimd) = 2/3 (25 – 3/2 (12) + ½ (–40) = –8.667 dBm = 136 µW

b) Ikinci derecli intermodulasyon neden ucuncu dereceli intermodulasyon kadar onemli degildir?

Ikinci derecli intermodulasyon isareti band disnda kaldigi icin filtre edilebilir. Fakat ucuncu dereceli intermodulasyon band icinde veya yakininda kaldigindan filtre edilemez?

5- Consider the following system.

BW=10 MHz G=20 dB BW=10 MHz G=—6 dB BW=1 KHz G= —3dB F=3 dB G= —3 dB F=6 dB G= —3dB OIP3=20 dBm OIP3=20 dBm a) Compute the cascaded noise figure.

F2−1 F 3 − 1 F 4 − 1 21− 21 − 41 − 21 − F=+ F 1 + + =+++2 + == 4.3 6.3 dB G1 GG 12 GGG 123 0.5 50 25 6.25

b) What S/N will be observed at the output when — 90 dBm signal is applied at the input.? (Ni=—144 dBm).

=− − = (Si N i ) 90 144 54dB = −=−= (SNoo ) ( SN i i ) F cas 546.3 47.7dB c) What is the IIP3 of the system?

OIP3= 11.1= 10.5 dBm IIP3= 2.5 dBm

2- Assume the following IF spectrum is applied to a bandpass filter with the given transfer function and the result is sensed by an amplifier:

a) Find the IIP3 of the amplifier such that the intermod product falling in the desired channel is 20 dB below the signal level.

b) Suppose a mixer with a voltage conversion gain equal to 10 dB and IIP3=500 mVp precedes the above circuit. Wht is the overall IIP3 of the system? (Neglect second order nonlinearities.)

3- When two signals with frequencies

a) From the graph, determine the RF input power level at which the 1 dB compression point is reached for the putput frequency w LO - w RF . PRF, 1 dB =12 dBm b) Show, on the graph, the 3 rd order interceetp point (IP3) for this mixer. Write the actual coordinates for the IP3 here: PRF = 13 dBm P IF = 7 dBm c) From curve A, what is the conversion loss (CL) of this mixer? CL = 6 dB. 4- Two tones are input into a nonlinear device with frequencies f1 and f2, f2 > f1. The lower frequency signal power is 6 dB higher than the higher frequency. What is the relative power ratio of the two third-order intermodulation products?

5- Assume the following IF spectrum is applied to a bandpass filter with the given transfer function and the result is sensed by an amplifier.

a) Find IIP3 of the amplifier such that the intermod product falling in the desired channel is 20 dB below the signal level.

(b) Suppose a mixer with voltage conversion gain equal to 10 dB and IIP3 = 500 mVp precedes the above circuit. What is the over all IIP3 of the system? (Neglect second-order nonlinearities.)

3- A nonlinear device is described by the folowing transfer function.

3- A 2 µV (RMS) carrier of 120 MHz is received at a 50 ohm antenna. The receiver consists of one radio frequency amplifier (10 dB gain), followed by a high side injection mixer (insertion loss 7 dB), followed by an IF filter (loss 2 dB), followed by a number of IF amplifiers (gain <= 25 dB), followed by the detector. The power received at the detector must be 3 dBm. fIF , = 10.7 MHz. a) Determine the power received by RF amplifier in dBm.

b) Determine the LO frequency.

c) What is the minimum number of IF amplifier stages necessary to meet the given specs. d) Draw a block diagram of the receiver (antenna through the detector). Show power level in dBm at each point.

e) Determine the image frequency. How can we eliminate receiver response to image frequency carrier.

29. An AM transmitter supplies 10kW of carrier power to a 50 Ω load. It operates at a carrier frequency of 1.2MHz and is 80% modulated by a 3kHz sine wave. (a) Sketch the signal in the frequency domain, with frequency and power scales. Show the power in dBW(If necessary, refer to the discussion of dBW in App. A.) (b) Calculate the total average power in the signal in watts and dBW. (c) Calculate the RMS voltage of the signal. (d) Calculate the peak voltage of the signal. Getting power in sidebands m2(0.8) 2 a) Psb= Pc = 10000 W = 1600 W 4 4 10000 W  Putting in dBW Pc ( dBW )= 10log  = 40 dBW at fc = 1.2MHz  1W  1600 W  PsbdBW( )10log=  = 32 dBW at fusb =+=+= fc fm 1.2 MHz 3 kHz 1.203 MHz and  1W   flsb=−= fc fm1.2 MHz − 3 kHz = 1.197 MHz

b) Ptotal=+ Pc 2 Psb = 10000 W +⋅ 21600 W = 13200 W 13200 W  Ptotal( dBW )= 10log  = 41.2 dBW  1W  Vtotal Ptotal R W V c) rms = ⋅= 13200 ⋅Ω= 50 812.4

d) Vcrms = PcR ⋅= 10000 W ⋅Ω= 50 707.1 V

EcVc=rms ⋅2 = 1000 V Emax= Ecm( 1 +=) 10001( += 0.8) 1800 WkW = 1.8 33. The spectrum analyzer display in Figure 3.16 represent the output of an AM transmitter. The analyzer has an input impedance of 50 Ω and is connected to athe transmitter output through a 60dB attenuator. Sketch the envelope of the signal in the time domain, as it would appear at the transmitter output terminals. Be sure to show both time and voltage scales.

Figure 3.16

Reference Level: -10dBm Vertical: 10 dB/division Center frequency: 21.200MHz Span: 5kHz/division

According Figure 3.16 5kHz fc = 2.21 MHz − = .21 1975 MHz 2 5kHz fm = = 5.2 kHz 2 Pc = −10 dBm −10 dB + 60 dB (attenuator to work at transmi tter output ter minal ) = 40 dBm = 10000 mW = 10 W Psb = −10 dBm − 20 dB + 60 dB (attenuator to work at transmi tter output ter minal ) = 30 dBm = 1000 W = 1W Psb 1W m = 2 = 2 = .0 6325 Pc 10 W Looking for Ec,Emax,Emin Ec = 2 ⋅ Pc ⋅ R = 2 ⋅10 W ⋅ 50 Ω = 623.31 V E max = Ec 1( + m) = 623.31 V 1( + .0 6325 ) = 6.51 E min = Ec 1( − m) = 623.31 V 1( − .0 6325 ) = 6.11 Let’s put signal definition: tf )( = 623.31 1( + .0 6325 ⋅ sin( 2π ⋅ 2500 t)) sin( 2π ⋅ 21197500 t) Plotting:

37. An AM signal appears across a 50 ohm load and has the following equation:

tv )( = 12 (1+ sin ( 566.12 ×10 3 t))sin ( 85.18 ×10 6 )Vt

(a) Sketch the envelope of this signal in the time domain. (b) Calculate the following quantities: (i) Carrier frequency (ii) Modulating frequency (iii) Total power (iv) Bandwidth

Checking data: Ec = 12 V m = 1 566.12 ×10 3 fm = = 2kHz 2π 85.18 ×10 6 fc = = 3MHz 2π Ec 2 ()12 V 2 Pc = = = 44.1 W 2R 2 ⋅ 50 Ω m 2 12 Psb = Pc = 44.1 W = 36.0 W 4 4 Pt = 44.1 W + 2 ⋅ 36.0 W = 15.2 As double sideband B = 2 fm = 2 ⋅ 2kHz = 4kHz E max = Ec 1( + m) = 12 V 1( + )1 = 24 V E min = Ec 1( − m) = 12 V 1( − )1 = 0V

Plotting Signal:

35. The circuit in Figure 4.22 can be used to indicate overmodulation. (a) Explain how it works. (b) Why will it not show very small amounts of overmodulation?

Figure 4.22

In normal operation modulated input drain current from main circuit through collector, but always current magnitude is small enough to drive current from diodes(LED). In overmodulation current is big enough to drive diodes into direct operation a turn on the LED. It is happen when current in the led is enough to put it on, that it is the cause that little overmodulation cannot be detected.

41. An AM radio station operates with a power of 10kW at a carrier frequency of 1 MHz. It is modulated at 75% by a 1kHz sine wave. The output is connected to a 50-ohm load. a) Sketch the transmitter output in the frequency domain, showing a frequency scale, and a power scale in watts. b) Sketch the transmitter output in the time domain (envelope display). Include time and voltage scales. First, it is necessary to get some data about signal fm=1 kHzTm , = 1sec, m fc = 1 MHzm , == 0.75, Pc 10000 W m2 ()0.75 2 Psb= Pc =10000 W = 1406 W 4 4 fusb=+= fc fm1 MHz + 1 kHz = 1.001 MHz flsb fc fm MHz kHz MHz =−=1 − 1 = 0.999 Ec=2 PcR =⋅ 210000 W ⋅Ω= 50 1000 V Emax= Ec (1 += m ) 1000 V (1 + 0.75) = 1750 V Emax= Ec (1 − m ) = 10 00V (1− 0.75) = 250 V ftV( )= 1000 (1 + 0.75sin(2π 1000 t ))sin(2 π 1000000 t )

With these values it is possible Sketch graphics.

31. A receiver uses low-side injection for the local oscillator, with an IF of 1750kHz. The local oscillator is operating at 15.750 MHz. (a) To what frequency is the receiver tuned? (b) What is the image frequency?

For low-side injection − = ⇒ = + = + = f sig f Lo f IF f sig f IF f Lo 75.1 MHz 75.15 MHz 5.17 MHz

− = ⇒ = − = − = f Lo fimage f IF f image f Lo f IF 75.15 MHz 75.1 MHz 14 MHz

38. The block diagram of Figure 5.25 shows a double-conversion receiver. (a) Does the first mixer use low- or high-side injection? (b) What is the second IF? (c) Suppose that the input signal frequency is changed to 17.000 MHz. What would the frequencies of the two local oscillators be now?

16.000MHz

First Mixer Second Mixer Detector Speaker

RF Amp Second First IF IF Amp AF Amp Amp 8.5MHz

First Local Oscillator Second Local 24.500MHz Oscillator 9.250MHz

> a) The first mixer have f Lo f sig then is high-side injection f = f b) sig 2 IF 1 = − = − = f IF 2 f Lo 2 f IF 1 25.9 MHz 5.8 MHz 750 KHz

= f sig 2 f IF 1 c) = + = + = f Lo 1 f IF 1 f sig 5.8 MHz 17 MHz 5.25 MHz

the frequency of the second oscillator is fixed.

35. The block diagram of an SSB transmitter is shown in Figure 6.27. The local oscillator frequency is higher than the frequency at which the SSB signal is generated, and the difference between the two frequencies is used at the output. (a) Choose a suitable frequency for the carrier oscillator if the transmitter is to produce a USB signal. (b) What should be the frequency of the local oscillator if the (suppressed) carrier frequency at the antenna is to be exactly 30 MHz? (c) Suppose that the transmitter is modulated by a single sine-wave tone at 1 kHz. It is operating with a PEP of 100 W into a 50 Ω load. Sketch the output in the time and frequency domains, showing all appropiate scales.

Figure 6.27

Carrier Oscillator Center 8MHz BW 2.7kHz

Local Mic Oscillator

a) Using limits of first filter: The second mixer executes a subtractive mixing with high-side injection. Then it takes lower band of signal at its input. Therefore, the first mixing should be lower- side band. = + 7.2 KHz = fup 8MHz .8 00135 MHz 2 7.2 KHz f = 8MHz − = .7 99865 MHz down 2

Using 300Hz from carrier to attenuate better the other sideband we can calculate values of carrier oscillator to LSB 300 Hz f = f + = .8 00135 MHz +150 Hz = .8 0015 MHz lsbc )( up 2

b) Calculating according formula in second mixer. f = f + f lo c1 co = + = For LSB flo .8 0015 MHz 30 MHz .38 0015 MHz

c) For spectrum, output signal only has one component at = + = + = f signal usb )( f co f m 30 MHz 1kHz 001.30 MHz with power of 100W

For envelope, getting some relevant data: = ⋅ ⋅ = ⋅ ⋅Ω = V p 2 R PEP 2 50 100 W 100 V 1 1 T = = = 1msec

f m 1kHz then tf )( = 100 V sin( 2π ⋅1000 t)sin( 2π ⋅ 30000000 t)

Note: Graphic application use limited number of points that make some approximation. Peak value must be 100V.

32) An FM receiver is double conversion with a first IF of 10.7 MHz and a second IF of 455 kHz. High-side injection is used in both mixers. The first local oscillator is a VFO, while the second is crystal-controlled. Three is one IF stage, one stage of IF amplification at the first IF, and three stages of combined IF amplification and limiting at the second IF. A Foster-Seeley discriminator is used as the detector. The receiver is tuned to a signal with a carrier frequency of 160 MHz. (a) Draw a block diagram for this receiver. (b) Calculate the Q that will be necessary in the input circuits to achieve a rejection of 50dB at the first image frequency, assuming that there are two tuned circuits with no coupling between them before the first mixer. [Hint: you will find the necessary equations in Chapter 5.] (c) If AGC were used with this receiver, to which stages could it be applied? (d) If AFC were necessary, from which stage would it be derived and to which stage would it be applied?

(a)

RF Amp 10.7 MHz IF1 Amp 455 kHz IF2 Amps and Limiter Discriminator

170.7 MHz 11.155 MHz 160 MHz

VFO

As two stages IF is produced by high-side injection. Then: f = f − f ⇒ f = f + f = 7.10 MHz +160 MHz = 170 7. MHz IF 1 osc 1 signal osc 1 IF 1 signal = − ⇒ = + = + = f IF 2 f osc 2 f IF 1 f osc 2 f IF 1 f IF 2 7.10 MHz 455 kHz 155.11 MHz

(b) Using formula of Image Rejection(IR) 2   2 f signal f image 1 f signal IR −1 IR = 1+ Q 2 ⋅  − −  ⇒ Q =   2  f image 1 f signal f image 1   f f   signal − image 1     f image 1 f signal  = + = + ⋅ = Getting fimage 1 f signal 2 f IF 1 160 MHz 2 7.10 MHz 181 4. MHz 50 2 − 20 − = IR 1 = 10 1 = Q 2 2 446 32.  f f  181 4. MHz 160 MHz   image 1 − signal   −     160 MHz 181 4. MHz   f signal fimage 1 

(c) AGC is supposed to be set to avoid overloading in AM signals. Due to FM discriminator is not affected a lot by change of amplitude. You can possible set in RF and IF’s stages, but it isn’t useful. In Discriminator, before AM detector, it could be useful set an AGC to reduce effect of frequency noise.

(d). AFC is used to keep discriminator within linear operation; it has affect the tunable part of receiver in this case VFO. Following figure shows the receiver with AFC.

RF Amp 10.7 MHz IF1 Amp 455 kHz IF2 Amps and Limiter Discriminator

170.7 MHz 11.155 MHz 160 MHz

VFO AFC