This document provides an exam for a radio electronics course, consisting of 5 questions testing knowledge of modulation schemes, transmitter and receiver design, wireless transmission systems, and linearity metrics. It includes the exam questions, solutions, and a list of important formulas for reference. The questions cover topics such as modulation constellations, bandwidth efficiency, transmitter frequency planning, noise figure calculation, image frequencies, output third order intercept point, and cascaded amplifier linearity.

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Integrated Circuits and Systems
http://www.ics.isy.liu.se/en/
TSEK02 – Radio Electronics
Tutorial 6
Exam solving
By Morteza Abbasi and Ameya Bihde 2013
2016-12-11 Updated by Ted Johansson (ted.johansson@liu.se)

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The exam has 100 points in total and your grade will be decided according to the following limits:
>85 5
>70 4
>55 3
(other limits may be used for other exams.)
Use the appendix containing the list of important formulae provided at the end of this exam paper.
Pay extra attention to units and conversions between dB and linear scale.
1. Compare QPSK, 16-QAM and 64-QAM modulation schemes in terms of their:
a) Constellation Diagrams. (6 p)
b) Bandwidth Efficiency. (4 p)
c) Peak power required in order to have the same symbol error probabilities. (10 p)
2. The heterodyne transmitter shown in the figure below uses a band pass filter (BPF) at the output
of the second up-conversion mixer to remove the undesired lower sideband. The band pass
filter characteristics are also shown below. If the carrier frequency is 1 GHz and the intermediate
frequency (IF) is 200 MHz.
a) Calculate the frequency of the local oscillator, FLO. (5 p)
b) Calculate the filter order required to achieve a sideband rejection of 50 dB. (10 p)
c) It turns out that the maximum filter order available in the laboratory is only 4. How should the
frequency plan of this transmitter now be redesigned? (10 p)
The bandwidth of the baseband can be neglected and assumed to be very narrow.
Baseband
IF
Mixer
RF Mixer
BPF
Antenna
FLO
0 dB
Attn = 20N dB/
decade
Centre
Frequency,Fc
Magnitude (dB)
Frequency (log scale)

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3. A point-to-point wireless transmission system should operate over 1 km at 5.8 GHz with 100
Mb/s and QPSK modulation. Raised cosine pulses with 𝛼 = 0.35 are used.
The transmitter can provide 22 dBm power and the antenna on both transmitter and receiver
sides have gains of 2 dB.
What should the noise figure of the receiver be if the minimum required SNR at its output is 8
dB? (Ignore any fading effect of the channel) (20 p)
4. A heterodyne receiver system operates over the frequency band of 15.3 - 15.7 GHz with four
channels and an IF of 1 GHz. Specify different LO frequencies for reception of different
channels and the corresponding image frequencies. What would the fractional bandwidth of the
channel selection filter be? (15 p)
5. A transmitter system is configured as shown below
a) Calculate the output third order intercept point of the transmitter (10 p)
After manufacturing it has been realized that this value is 2 dB less than the requirements. In
order to avoid any re-design, an additional linear amplifier with 20 dB of gain is added to the
output and the input signal will be reduced accordingly.
b) How much should the OIP3 of the amplifier be so that the transmitter meets the linearity
requirements? (10 p)
Modulated
Signal
gain= 10 dB
OIP3=10 dBm
Loss=0.5 dB gain= 25 dB
OIP3=28 dBm
gain= 6 dB
OIP3=5 dBm

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Solutions
The exact numerical results are very sensitive to rounding errors and some deviation in the
numbers is tolerated and still give full points for the answer! The important is that the examiner
is able to follow your calculations and thoughts!
1.
Check Tutorial 1, Problem 1.3
2.
Similar (but not identical) to Tutorial 1, Problem 1.5
a)
fC = fLO +/- fIF
fC = 1 GHz, fIF = 200 MHz
We filter the lower sideband and use the upper sideband.
fLO = 800 MHz.
b)
Image is 2* fIF away at 600 MHz.
The numbers of decades is ABS((fC - 2 fIF) / fC)
This is ABS(log(600 MHz/1 GHz)) decades = 0.22 decades.
The required attenuation is 50 dB = 20N *0.22 dB => N=11.27 -> 12.
c)
Max order is 4. We keep fC but change fIF.
ABS(log((fC - 2 fIF) / fC)) * 20 * 4 = 50 => fIF = 381 MHz.
3.
We use Friis' transmission formula.
λ =
C
f
=
3×108
5.8×109 = 0.05172m
Convert transmit power and antenna gains into linear scale.
The power reaching the receiver input is PRX = PTX
Gant
2
×λ2
(4πR)2 = 6.7448 × 10−12
W
With QPSK modulation each symbol represents two bits so the symbol rate is Rs=Rb/2=50 Msym/sec
Bandwidth of the signal is Rs(1+) = 50M(1+0.35) = 67.5 MHz.
Psen (Pmin) = -174 dBm/Hz + 10log(B) + NF + SNRout.
B = 67.5E6, SNRout = 8 dB, Pmin = 6.7448E-12 W = -81.71 dBm
=> NF = 5.9967 dB = 6.0 dB.

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4.
The frequency band is divided into 4 channels. Bandwidth of each channel is
B =(15.7-15.3)/4 = 0.1 GHz
Center frequencies of these channels are therefore [15.35 GHz, 15.45 GHz, 15.55 GHz, 15.65 GHz].
With 1 GHz IF, LO frequencies could be either [16.35 GHz, 16.45 GHz, 16.55 GHz, 16.65 GHz] or
[14.35 GHz, 14.45 GHz, 14.55 GHz, 14.65 GHz] and image frequencies are either [17.35 GHz, 17.45
GHz, 17.55 GHz, 17.65 GHz] or [13.35 GHz, 13.45 GHz, 13.55 GHz, 13.65 GHz], respectively.
Fractional bandwidth of the channel select filter should be
B
f
=
0.1
1
= 10 %.
5.
a) The output IP3 of the transmitter is calculated to 357 mW = 25.5 dBm.
(No OIP3 given for the filter with loss=0.5 dB, therefore assume OIP3 of this block to be very large, use
e.g. OIP3 = 100 dBm in the calculations).
Here is a calculation in matlab, so it is easy to track where the values come from and less risk of
rounding errors:
b) The required OIP3 has been 27.5 dBm = 562 mW.
The final OIP3 with the new amplifier is
1
OIP3new
=
1
Gamp×OIP3old
+
1
OIP3amp
so OIP3 of the amplifier
should be OIP3amp = 1/ (
1
562
−
1
100×357
) = 570 mW = 27.6 dBm.
(Did not use the exact numerical values from a) in this calculation, just three digits as shown.)

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Important Note: Always watch out for the scale. Check whether you are in dB scale or the linear scale.
This is a very common mistake.
List of Important Formulae
1. Shannon’s Channel Capacity Theorem
𝐶 = 𝐵 × log2(1 + 𝑆𝑁𝑅) = 𝐵 × log2 (1 +
𝑆
𝑛0 × 𝐵
) [𝑏
𝑠
⁄ ]
n0 is the noise power spectral density in W/Hz, S is the signal power in W, B is the bandwidth,
SNR is NOT in dB scale. Also note the log2 which is not the common log10.
2. Bandwidth of a signal shaped by a raised cosine pulse filter is
1+𝛼
𝑇𝑏
α is the roll-off factor, Tb is the original pulse period.
3. Boltzmann’s Constant, k = 1.38 x 10-23
J/K.
4. Use a room temperature of 27 °C = 300 K whenever temperature is not specified.
5. Thermal noise power spectral density, PSD=kT. At T=300 K, PSD is -174 dBm/Hz. The PSD is
independent of the resistor value. This is true only when the source resistor and the load
resistances are matched.
6. Thermal noise power in a bandwidth B: PRS = kTB.
In dB scale at 300 K, the total thermal noise power PRS|dB = 10log(kTB) = 10log(kT)+10logB
=> PRS|dB = -174 dBm/Hz + 10logB
7. Noise Factor [not in dB]
out
in
SNR
SNR
NF 
Noise Figure [dB] 








out
in
dB
SNR
SNR
NF log
10 = SNRin|dB -SNRout|dB
8. Noise figure of a passive lossy component is equal to its loss: NF=L.
9. Effective noise figure of cascaded stages.
A B C
GA, NFA GB, NFB GC, NFC
B
A
C
A
B
A
total
G
G
NF
G
NF
NF
NF
1
1 



 .
This is called Friis’ equation. This equation is not in dB.
10. IP3 = P1db + 9.6.
in dBm and valid for both input and output referred quantities

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11. Output IP3 of a component can also be calculated from the two-tone test:
𝑂𝐼𝑃3 [𝑑𝐵𝑚] = 𝑃1[𝑑𝐵𝑚] +
∆𝑃[𝑑𝐵𝑐]
2
where P1 is the power of each of the main tones, ∆𝑃 is the power difference between the two
tones and the distortion tones.
12. IP3 = P + ΔP/2.
in dBm and valid for both input and output referred quantities.
P is the input/output power in each of the main tones, ΔP is the power difference between the
main tones and the distortion tones
13. IP3 of cascaded stages:
A B C
Effective IIP3 (in W, not in dBm/dB)
C
B
A
B
A
A
total IIP
G
G
IIP
G
IIP
IIP 3
3
3
1
3
1


 , where G is the gain.
If referred to the output, OIP3 becomes
C
B
C
A
C
B
total OIP
OIP
G
OIP
G
G
OIP 3
1
3
1
3
1
3
1





14. At 300 K, the power required at the receiver input in dBm for a given output SNR in a bandwidth
B is given by Pin|dBm =-174 dBm/Hz+10log(B)+NFdB +SNRout|dB. .
15. Dynamic Range Linear (referenced to input) in dB: DRL= P1dB(referenced to input) - Psen.
16. Spurious Free Dynamic Range, SFDR (referenced to input) in dB:
SFDR =
2 PIIP3 +174 dBm/Hz - NF -10logB
( )
3
- SNRmin
This formula assumes that the input noise is thermal at 300 K.
17. After propagation through an ideal channel of R meters, the received power level is given by
𝑃𝑟𝑒𝑐𝑒𝑖𝑣𝑒 = 𝑃𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡 × 𝐺𝑡 × 𝐺𝑟 ×
𝜆2
(4𝜋𝑅)2
𝐺𝑅 and 𝐺𝑇 are receive and transmit antenna gains and 𝜆 is the wavelength given by 𝜆 =
𝑐
𝑓
, where
c=3*108
m/s.