35

Magnetrons are cheap, reliable, pretty efficient (65% or so- and they tolerate high temperatures so heat sinking is easy) and made with mature technology. They are also reasonably tolerant of VSWR issues (if the user does not put a proper load in the oven, for example). They don't really allow the frequency to change much without expensive mechanical tuning ...


19

The specific formula applies only for a first order RC low pass filter. This is derived from its frequency response: $$H(j\omega)=\frac{1}{1+j\omega RC}$$ The cutoff frequency is defined as the frequency where the amplitude of \$H(j\omega)\$ is \$1\over\sqrt2\$ times the DC amplitude (approximately -3dB, half power point). $$|H(j\omega_c)|=\frac{1}{\sqrt{...


16

For a simple RC low pass filter, cut-off (3dB point) is defined as when the resistance is the same magnitude as the capacitive reactance: - \$R = \dfrac{1}{2\pi f C}\$ It's a simple math trick to say: - \$f = \dfrac{1}{2\pi R C}\$


13

The R&S instruments are equipped with female test port adapters. Edit: archived pdf You can replace the N type with a more suitable type for higher frequencies.


13

H(w) =1/ (square root of 2) That sounds a bit confusing, we usually refer to the cutoff point as the -3 dB point. That is the same though, -3 dB is half the power. Let me explain: take your \$H(\omega) = \frac{1}{\sqrt2}\$ That means that at that \$\omega\$ the voltage is divided by \$\sqrt2\$, if this voltage is applied across a (load) resistor at the ...


13

RLC Passband and Stopband Your question doesn't explicitly take note of the fact that there are four distinct RLC filters that are either band-pass or band-stop. But I'd like to list them for those interested in a slower pace: simulate this circuit – Schematic created using CircuitLab The diagrams with the capacitor and inductor in parallel are called ...


12

EDIT: Thanks to hryghr I see that the starting assumptions were incorrect. The transfer function magnitude can't be found that simply. It is more than ten years since I considered my skills sharp on this topic, and knives don't get sharper in the drawer! But I can't have that I posted something formally incorrect, so here goes attempt #2: I will derive the ...


12

The domestic microwave oven needs high power to cook the meal and high frequency to excite the water molecules. What is not needed is high stability because the water energy absorption spectrum is broad. (1, 2) The magnetron does this cheaply. The low price and low duty cycle of the domestic microwave means that they should last for many years despite ...


8

The 3-dB cutoff is just one commonly used way to describe a filter. For some applications you might want to specify the 10-dB cutoff or 60-dB cutoff instead. It is convention that if someone says "cutoff frequency" without being more specific, they are talking about the 3-dB cutoff. You should think of the 3-dB cutoff as the frequency where the filter begins ...


8

What is the purpose of this simple 1 capacitor 2 resistor [high] pass filter? The extra resistor gives you an extra degree of freedom. For the standard RC high-pass filter in the first schematic, the transfer function is $$H(j\omega) = \frac{j\omega R_1C}{1 + j\omega R_1C} $$ So, the asymptotic high frequency gain is 1 and the corner frequency is \$f_c=...


8

It's not 3dB absolute, it's 3dB down from the peak, or some sort of nominal attenuation. So in your case, where the passband is -3dB, 3dB down is at -6dB. Note that some filters (e.g. Chebychev) have significant passband ripple; if this exceeds 3dB then the "3dB down" figure loses meaning. In that case, or just if it's what matters to the system ...


7

A lot of people confuse natural frequency with cut off frequency. The natural frequency is the frequency the system wants to oscillate at. The cut off frequency (or -3dB freq) is just when the transfer function has a magnitude of 0.707 If the two poles of the filter are not close together, the 2nd order canonical terms like the natural frequency and the ...


7

First, that's a high-pass filter. Secondly, if you do not load Vout (on the second circuit), then I think you can see that the series resistance R1 + R2 is equivalent to R1 (on the first circuit), so you can find the -3dB frequency. The output voltage at high frequencies (second circuit) will be \$ V_{OUT} = V_{IN} \cdot \$\$R_2 \over R_1 + R2\$, and the ...


7

Not that I'm aware of, call it a filtered voltage divider :) \$fc=\frac{(R_1+R_2)\sqrt{(10^{3/10}-1)}}{2\pi*R_1*R_2*C_1}\approx \frac{(R_1+R_2)}{2\pi*R_1*R_2*C_1} \$ \$R_1//R_2//C_1 \$ There are other considerations to take into account. In your case for instance, the conversion speed (the time it takes to charge the ADC's internal sampling capacitor) ...


6

For a voltage transformer: - If the applied voltage were very low in frequency i.e. 0.1 Hz, the primary current taken (if the voltage were 120 VAC or 230 VAC) would be several tens of amps and the winding would fry and burn. Nonetheless, what about trying it at 1 VAC? Now, the current (remember we're talking 0.1 Hz) is a couple of hundred milliamps (that's ...


6

Spehro gave you the answer, now let me tell you why you could (should?) have known that by looking at the circuit. Two resistors in series are indistinguishable from one resistor with the sum of the two resistances. (I hope you knew this?) Hence when we take Vout in the second circuit from the R1/C1 junction, we have exactly the same circuit as the first (...


6

It is not correct that for "all filters" the corner or cut-off frequency is defined by the "-3dB point" (magnitude 3 dB down with respect to the maximum). This is only the case for all first-order low- and highpass responses as well as 2nd-order bandpass filters, and for higher-order filters with Butterworth characteristics. For all ...


5

Non-superconducting current transformers have a rolloff below a certain frequency, so the output for a given current simply goes down as the frequency decreases. It's not core saturation nor is it directly related to the inductance- this is a linear effect. The -3dB rolloff frequency is determined from L/R where L is the inductance of the transformer ...


5

From the simulation, it looks like it fc=500Hz or somewhere around there. Your simulation looks more like 15 kHz (3 dB point) and the reason is because you haven't used enough resolution in your AC analysis and the graph is inaccurately interpolating between points that are too far apart: - Use more resolution in your AC analysis like in this simulation ...


5

Well, mathematically speaking we can write: $$\mathcal{H}\left(\text{s}\right)=\frac{\text{sL}}{\text{sL}+\text{R}}\tag1$$ Using \$\text{s}=\text{j}\omega\$, we get: $$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\left|\frac{\text{j}\omega\text{L}}{\text{j}\omega\text{L}+\text{R}}\right|=\frac{\left|\text{j}\omega\text{L}\right|}{\left|\...


5

Back of the envelope calculation... 20k/4250 = about 4 It's a second order filter, so at a frequency 4x above its -3dB point it should attenuate the signal by 4 squared, or about 16. Output signal is about 0.3Vpp. Input signal is 5Vpp. 5/0.3 = about 16. So it delivers the correct attenuation. If you want more attenuation, use a lower -3dB point, or a higher ...


5

It seems to be your task to figure out how to do this, and we don't solve homework for people. But your actual question here seems to be: I don't understand why I should find the method to determine cutoff-frequency. The real life of an engineer involves not only designing things, but also verifying that your design works as intended (or more often, figuring ...


5

The reason the "gain" is above zero is that it's not gain, it's voltage expressed in dBV and you're putting 5 VAC into the circuit. Change your source AC amplitude to 1 and try it.


4

This is a filter with a resistor divider. It's NOT a lowpass filter, it's a highpass filter, often used to block DC. The divider is presumably used to scale the input to an appropriate range for whatever is connected to Vout. Assuming the load on Vout is infinite impedance, the corner frequency should be - $$ f = \frac{1}{2 \pi (R1 + R2) C}$$ At ...


4

For a lowpass with order n=8 you can expect a magnitude drop (far above the cut-off frequency) of 48dB/Octacve. I think, the presented curve does show such a slope - however, only approximately. Why do you expect a damping of 40 dB at 2 kHz? More than that, Bessel filters are optimized with respect to their phase response (linear). The price paid for this ...


4

Solving a simple 2nd-order circuit like this one requires a few lines obtained by inspecting the circuit. This is how Fast Analytical Circuits Techniques or FACTs described in "Linear Circuit Transfer Functions: an introduction to FACTs" work. Ok, start with \$s=0\$: remove all caps. The dc gain \$H_0\$ is 1. The denominator is obtained by setting the input ...


4

In my opinion this 'cut-off' frequency is not defined as the -3dB point. The real transfer function is: $$ H(s)=\frac{1}{s^2R_1R_2C_1C_2+s(R_1C_1+R_1C_2+R_2C_2)+1}$$ The 'common form' of a second order element in control theory is $$W(s)=\frac{1}{\frac{s^2}{\omega_n^2}+2\frac{\xi}{\omega_n}s+1}$$, where \$\xi\$ is the damping coefficient and \$\omega_n\$ is ...


4

The browser doesn't like that link to the datasheet, so I couldn't read it. Provide the link to just the PDF file, not a page with all kinds of fluff around it. In any case, the reason for the low pass filter is that the motor current can have short term spikes and other noise, but what you care about is more of a recent "average", or more precisely, you ...


4

I show you how to obtain the 3dB cut-off frequency for the low pass filter \$G_1(s)\$. You can calculate the cut-off frequencies of the band pass filter \$G_2(s)\$ in a similar way, as long as you know that its maximum magnitude is attained at \$\omega=\sqrt{b}\$, as pointed out in a comment by robert bristow-johnson. The latter fact can be derived by ...


4

For this type of a circuit the cutoff frequency is not at Vin/Vin = 0.707. Because the R1 and R2 form a voltage divider, and "mid-band gain" (maximum of Vout) is equal to R2/(R1 + R2) = 0.9[V/V], therefore the cutoff frequency is at 0.9*0.707 = 0.636(the frequency at where the magnitude of Vout is 3dB less than the maximum of Vout). And this is why right ...


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