Question

Problem #6: You are tasked to design a 10 GHz microstrip filter using the Rogers RT5870, which has a dielectric constant of 2.33. The substrate has the following thickness (d) options, 0.254 mm, and 1.575 mm. a. For both thickness options, find the line width (W) to obtain a 50 line. (hints assume W/d <2). b. Compute the higher-order resonance frequencies for both thicknesses. Which option gives a higher operation frequency?

Answer 1

As perthe question, we need to design a 10 GHz microstrip filterwhich has a dielectric constant of 2.33. The substrate has the followingthickness 0.254 mm and 1.575 mm. We needto find out the width (w) of the line in orderto obtain 50 ohm single ended impe dance. Microstripis a special kind of transmission line which can be easily fabricated. It can easily generate microwave-freque ncy. The conducting strip is separated from the ground plane by a specialized dielectric constant. The main advantage of these types of systemisthat it is less expensivethan traditional wave guide material. However, in this technology, the loss is a bit higher. The generalized width of the microstrip can be calculated as by the followingformula 7.48 h 1.25* d w = E, + 1.41 еxp (2 * V 87 Where Z is the characteristics impedance, his the diele ctric thickness, d is track thickness and e, is the relative dielectricthickness. Now as per the question, the dielectric constant of the material is given, from this information we need to calculate the effective resistance (e,) which can be evaluated as by the following formula E1 + 2 1 E-1 E = 1+12( w Let us assume that= 2. Then by the inserting the value of s, the effective resist an ce (e,) is Er1.80

Let us assume that= 2.Then by the inserting the val ue of s, the effe ctive resistan ce (s,) is E = 1.80 Now insertingthe avlues obtained from the previous equation, we can get the value of w such that 7.48 0.254 1.25 (0.254/2) w = 1.80 1.41 exp (50 87 In this calculation, we have assume that h=0.254. Then the value of w will be W 0.55 mm Now the same analysis has been performed for h=1.575 mm. Afterinserting the value we obtain the value of w will be

7.48 1.575 1.25 (1.575/2) W = V1.801.41 exp (50 87 Then the value ofw will be; W 3.22 mm. b. Afterthat we need to calcul ate the resonance frequency for both cal culated thickness. The formula of the generalized resonant frequency of a microstrop can be cal cul at e d as т 2 + 2VEW fmn Where cis the velocity of the em wave in free space, m and n are the positive integers. Now the fundamental TM10 can be cal cul ated as for w-0.55 mm and d-0.254mm 3 * 1011 =203 GHz fio = 2+0.55*V1.8 2wVET The same approach can be followed forw=3.22 mm and d-1.575mm. In that case, the TM10 fundamental resonant frequency can be cal culat e d as *101134 GHz fio 2+3.22 1.8 2w ET From the analysis, we can conclude that the first option will provide the higher range of fundame ntal resonant frequency