Hartley vs Weaver Image-Reject Receivers Compared
Both receivers cancel the image by combining quadrature I and Q paths, but the rejection is always finite. For small errors IRR ≈ 4/(ε² + θ²), so a 1% gain error with a 0.01 rad phase error gives about 43 dB.
For practitioners
A heterodyne receiver maps two RF frequencies onto the same IF: the wanted signal at f_LO + f_IF and the image at f_LO − f_IF. Once the mixer has run, nothing can separate the two. The classic fix is a preselect filter ahead of the mixer, but that gets hard when the IF is low and the image sits close to the wanted channel. Image-reject receivers solve the problem with phase instead of frequency selectivity. The Hartley and Weaver architectures are the two standard ways to do this.
Why phase can cancel what a filter cannot
Both receivers split the signal into I and Q paths. They do this by mixing with LO phases 90 degrees apart. In one path the wanted signal and the image come out with the same sign. In the other path they come out with opposite signs. Combining the two paths correctly makes the wanted terms add and the image terms subtract.
This only works when the two paths match. The image cancels only when its two copies have equal amplitude and are exactly 180° apart at the combiner. Every real limit of these receivers comes back to that condition.
The Hartley receiver: quadrature mixers plus a 90° shifter
Write the input as a wanted tone A·cos(ω_RF t) plus an image B·cos(ω_IM t), where ω_RF = ω_LO + ω_IF and ω_IM = ω_LO − ω_IF. The I mixer is driven by LO (0°) = cos(ω_LO t) and the Q mixer by LO (−90°) = sin(ω_LO t). After each mixer, a low-pass filter removes the sum-frequency products and leaves:
In the I path the wanted signal and the image share a sign. In the Q path their signs are opposite. Next the I path goes through a 90° phase shifter. With a +90° shift, cos becomes −sin:
The B terms cancel and the wanted signal remains. If you swap addition for subtraction, or flip the sign of the shift, the receiver keeps the image side and rejects the upper side. That is how one structure can select either sideband.
The weak point is the shifter, labelled in the top diagram. It has to give an accurate 90° across the whole IF bandwidth, not just at one frequency. A common textbook implementation is an RC–CR pair. It gives exactly 90° between its outputs at every frequency, but the outputs have equal amplitude only at f = 1/(2πRC). Move away from that frequency and the network adds a gain error. Component tolerance shifts the balance point as well. The result is that a Hartley receiver's image rejection depends on how well the shifter holds gain and phase across the channel.
The Weaver receiver: a second quadrature mixer pair
The Weaver architecture keeps the first stage: quadrature mixers driven by LO1 at 0° and 90°, then low-pass filters. Instead of a phase shifter, each path feeds a second mixer, and the two second mixers are driven by a quadrature LO2. A summer follows, as the bottom diagram shows.
The second quadrature mixing produces the same relative sign flip that the shifter produced in the Hartley design. After the second stage, the wanted components from the two paths are in phase and the image components are in antiphase, so the summer cancels the image. The final IF is set by the combination of LO1 and LO2.
There are two practical consequences. First, there is no frequency-dependent RC network, so a wide IF bandwidth no longer ruins the 90° accuracy. The quadrature accuracy now depends on how well LO2 holds its quadrature phase. Second, the second mixing stage has its own image band. The low-pass filters between the stages must suppress signals in that band before they reach the LO2 mixers.
How gain and phase mismatch limit image rejection
Neither structure reaches infinite image rejection. Let ε be the fractional gain error between the I and Q paths and θ the phase error in radians. For small mismatch, the image rejection ratio (a power ratio) is approximately:
This has the same form as the IQ imbalance result, because it is the same physics. Gain and phase errors enter as a sum of squares, so the larger error dominates. In the Hartley receiver, ε and θ include the mixer and filter mismatch plus the shifter error. In the Weaver receiver, the mismatches of both mixing stages contribute to the total error.
On the curve for ε = θ, the formula reduces to IRR = 2/ε². That gives 63.0 dB at 0.001, 43.0 dB at 0.01 (IRR = 20000), 29.0 dB at 0.05 and 23.0 dB at 0.1 (IRR = 200). Each tenfold increase in mismatch costs 20 dB.
Worked example: 2% gain error and 1° phase error
Take ε = 0.02 and θ = 1°.
- θ in radians: 1 × π/180 = 0.01745 rad
- ε² = 0.000400, θ² = 0.000305, sum = 0.000705
- IRR = 4 / 0.000705 ≈ 5677
- IRR_dB = 10·log10(5677) ≈ 37.5 dB
Now suppose this is a Weaver receiver, and its second stage has the same 2% and 1° errors. Assume, for illustration, that the two stages' errors have the same sign and add directly. That is the worst case. Then ε = 0.04 and θ = 2° = 0.0349 rad. The sum of squares is 0.0016 + 0.00122 = 0.00282, so IRR ≈ 1419, or about 31.5 dB. Doubling every error costs 6 dB.
Choosing between them
Hartley uses fewer mixers and one LO. It suits narrow IF bandwidths, where a passive shifter can stay balanced across the channel. Weaver avoids the wideband shifter and handles wider IF bandwidths better. The cost is a second LO, a second mixer pair, and a secondary image that the inter-stage filters must handle. In both designs, the achievable rejection follows from IRR ≈ 4/(ε² + θ²) applied to the total path mismatch.
In plain words · for students and beginners
Image-reject receivers in plain words
An everyday analogy, and where it breaks down
Picture two friends carrying the same two messages, one good and one unwanted. One friend carries both messages as written. The other friend carries the good message as written and the unwanted one with a minus sign in front. When you add what they bring, the unwanted message disappears and the good one doubles.
That is the core trick. The image is an unwanted signal on the other side of the LO (local oscillator, the receiver's internal reference tone). After mixing, the image lands on exactly the same frequency as the wanted signal.
The analogy breaks down in one place. Radio signals are waves, not plain numbers. The "minus sign" comes from timing, or phase (where the wave is in its cycle). The two paths are called I and Q (in-phase and quadrature, meaning 90 degrees apart). If one friend's copy is a bit louder, or a bit late, the unwanted message does not fully disappear. A thin leftover remains.
Hartley and Weaver in one line each
- Hartley: mix with two LO copies 90° apart, filter, delay one path by a further 90° with a phase shifter (a circuit that shifts a wave's timing), then add.
- Weaver: same first step, but replace the phase shifter with a second pair of mixers driven by a second LO.
The worked example in simple steps
Say the two paths differ in strength by 2%, and in timing by 1 degree.
- Write the gain error as a fraction: ε = 0.02.
- Convert 1 degree to radians (the angle unit the formula uses): 1 × 3.1416 / 180 ≈ 0.01745.
- Square both and add: 0.0004 + 0.000305 ≈ 0.000705.
- Divide 4 by that: 4 / 0.000705 ≈ 5677. This is the image rejection ratio, how many times weaker the image becomes.
- Convert to dB (decibels, a log scale for ratios): 10 × log10(5677) ≈ 37.5 dB.
So the image comes out about 5677 times weaker than it went in. It is reduced, not removed.
Mistakes beginners make
- Expecting perfect cancellation. The rejection is always finite. Even 0.1% errors in both gain and phase stop at about 63 dB.
- Mixing degrees and radians. Put 1 into the formula instead of 0.01745 and the answer is wildly wrong.
- Forgetting the dB conversion. IRR from the formula is a power ratio. You need 10·log10 to get dB.
- Thinking Weaver has no errors. It has two mixing stages, and both add their mismatch to the total.
Remember
- Both receivers cancel the image by combining I and Q paths that are 90° apart.
- Hartley needs an accurate 90° shifter over the IF bandwidth. Weaver replaces it with a second quadrature mixer pair and a second LO.
- IRR ≈ 4/(ε² + θ²). For 1% gain error and 0.01 rad phase error, that is about 43 dB.
Related pages on RF View
Frequently asked questions
What is the difference between a Hartley and a Weaver image-reject receiver?
Both mix the RF signal with quadrature LO signals to make I and Q paths. Hartley then shifts one path by 90 degrees and sums. Weaver uses a second pair of quadrature mixers, driven by a second LO, in place of the phase shifter.
How much image rejection can a Hartley or Weaver receiver achieve?
It is limited by gain and phase mismatch between the I and Q paths. For small errors IRR ≈ 4/(ε² + θ²). With equal 0.01 gain error and 0.01 rad phase error this works out to 20000, or about 43 dB.
Does the Weaver architecture remove the need for an accurate 90-degree phase shifter?
Yes. It replaces the shifter with a second quadrature mixer pair. The second stage still needs matched paths, though, and its gain and phase errors add to those of the first stage.
Why is image rejection never infinite in a real receiver?
The cancellation relies on two image components of exactly equal amplitude and exactly opposite phase. Any gain or phase error leaves part of the image behind, so the rejection ratio is finite.