Solution manual for microwave circuit design a practical approach using ADS by yeom

Solution A post-wall waveguide also known as substrate integrated waveguide SIW or a laminated waveguide is a synthetic rectangular electromagnetic waveguide formed in a dielectric subs

Note:

When Figures in the text are cited, it is denoted by Figure and figures in this End-of

chapter solutions are denoted by Figure

CHAPTER 1 PROBLEMS

1.1 Waveguide generally has lower line loss than microstrip SIW(Substrate Integrated Waveguide)

can be regarded as the planar version of waveguide How SIW is configured using substrate?

Solution

A post-wall waveguide (also known as substrate integrated waveguide (SIW) or a laminated waveguide) is a synthetic rectangular electromagnetic waveguide formed in a dielectric substrate by densely arraying metallized posts or via-holes which connect the upper and lower metal plates of the substrate The waveguide can be easily fabricated with low-cost and mass-production using through-hole techniques where the post walls consists of via fences The post-wall waveguide is known to have similar guided wave and mode characteristics to the conventional rectangular waveguide with equivalent guided wavelength The structure is shown below Since SIW provides low loss, filters using SIW have studied by many researchers

Figure 1.1 The structure of SIW

1.2 Find the microwave integrated circuit example built using LTCC in the web-site

Solution

In the web site http://www barryltcc.com/, there is a TR (Transmit and Receive) module for radar

1.3 How the ALC in Figure 1.7 is constructed?

Solution

Figure 1.2 shows a block diagram for ALC Usually PA module has the terminal for output power control If not, a driver amplifier should be designed to control the output power level A small portion of output power is sampled through a directional coupler as shown in Fig 1.2 and the RF power level is detected through a diode detector The reference signal is a variable DC voltage

source Sometimes when the variable voltage source is difficult, it is generated using a lowpass filter PWM (Pulse width modulation) signal generated from the controller The average DC voltage corresponds to the desired power level Since it is a negative feedback loop, the detector voltage is equal to the reference voltage at the steady state Thus the output power level can be controlled by changing the reference voltage

Figure 1.2 ALC block diagram

1.4 Refer to SA605 which is used to demodulate the FM signal Explain its block diagram

Solution

FM demodulation in SA605 is based on quadrature FM demodulator shown in Fig 1.3 In coming

FM signal is split in two paths; one is directly applied to the mixer and the other is coupled to the resonant circuit centered at the FM carrier frequency by a high reactance capacitor As a result, the signal coupled to the high reactance capacitor suffers a 90 phase shift The FM modulated signal suffers additional phase shift proportional to the resonator slope and AM modulated waveform proportional to FM modulation appears at V2 As a result of mixing demodulated FM waveform appears

Figure 1.3 Quadrature FM demodulator

1.5 Refer the web-site of PLL IC Explain the synthesizer data bus shown in Fig 1.7

Solution

Refer to Analog Devices, RF PLL Frequency Synthesizers ADF4110/4111/4112/4113, August,

2012

1.6 How the PLL can be modulated? Explain how to set the PLL loop bandwidth

Solution

For FM signal generation at the VCO output, the modulation signal may be directly injected to VCO tuning input The fast varying signal than PLL loop bandwidth can modulate VCO successfully, however the slow varying one will be compensated by the PLL and does not present the modulated output

Figure 1.4 FM modulation using PLL

CHAPTER 2 PROBLEMS

2.1 A company A fabricates a thin film resistor whose sheet resistivity is 50 ohm/square and in the

case of a company B, an identical thin film resistor has a sheet resistivity of 100 ohm/square What

is the difference between the processes of the two companies? Furthermore, given that the material’s volume resistivity is ρ and its thickness is t, find its sheet resistivity

Solution

The thickness of the resistor material of the company A is two times thicker than that of the company B

The resistance R with a volume resistivity of ρ and a thickness of t is

S

ρ

Thus

S

R t

ρ

=