深度学习中感受野尺寸、目标尺寸及VGGNet 500×500输入感受野计算问询

Hey there! Let’s break this down clearly, starting with the core concepts and then validating your VGGNet receptive field calculations.

Receptive Field Size #

The receptive field size of a neuron in a CNN refers to the size of the region in the original input image that this neuron "sees" to produce its output. In simpler terms, it’s how much of the input image contributes to a single feature map pixel in that layer. A larger value means the neuron can capture broader contextual information from the input.

Target Size #

When we talk about "target size" in this context, it usually refers to two things:

• The actual pixel dimensions of the object you’re trying to detect/recognize in the input image (e.g., a cat that’s 200×150 pixels in your 500×500 input).
• It can also refer to the spatial dimensions of the feature map output by a specific layer (this is the Output size in your calculation table).

First, let’s recap the standard recursive formula for calculating receptive fields:
Current Layer RF Size = Previous Layer RF Size + (Current Layer Kernel Size - 1) × Cumulative Stride Up to Previous Layer
Note: Activation layers like ReLU don’t change receptive field size, output size, or stride—they only apply element-wise transformations, so they preserve all spatial properties.

We’ll start with the input layer (500×500 image), where the initial RF size is 1 and cumulative stride is 1.

Verified Layers #

Let’s go through each layer you listed:

• conv1: Kernel size 3, stride 1
  RF = 1 + (3-1)×1 = 3; Output size = (500 - 3 + 2×1)/1 + 1 = 500 (thanks to same padding in VGG). Your result is correct.
• relu1_1: As an activation layer, it keeps RF size, output size, and stride identical to conv1. Your result (RF=3, Output size=500, Stride=1) is right.
• conv1_2: Kernel size 3, stride 1
  RF = 3 + (3-1)×1 = 5; Output size remains 500 (same padding). Your result is correct.
• relu1_2: Again, activation layer preserves all values. RF=5, Output size=500, Stride=1 is accurate.
• pool1: Pooling kernel size 2, stride 2
  RF = 5 + (2-1)×1 = 6; Output size = (500 - 2)/2 + 1 = 250; cumulative stride becomes 1×2=2. Your result is perfect.

For the Unfinished conv2_1 Layer (for reference) #

If we continue to conv2_1 (VGG uses kernel size 3, stride 1, same padding):

• RF = 6 + (3-1)×2 = 10
• Output size stays 250 (same padding)
• Stride remains 2 (since conv stride is 1, cumulative stride doesn’t change)

Overall, all the calculations you provided are 100% accurate—they align perfectly with VGGNet’s architecture and standard receptive field calculation rules.

内容的提问来源于stack exchange，提问作者batuman